GMAT Quant

As if solving inequalities wasn’t already hard enough, sometimes the way a GMAT question is framed will make us wonder which answer option to choose, even after we have already solved solved the problem.

Let’s look at three different question formats today to understand the difference between them:

We have two linked inequalities here. One is |-x/3 + 1| < 2 and the other is the correct answer choice. We need to think about how the two are related.

We are given that |-x/3 + 1| < 2. So we know that x satisfies this inequality. That will give us the universe which is relevant to us. x will take one of those values only. So let’s solve this inequality. (We will not focus on how to solve the inequality in this post – it has already been discussed here. We will just quickly show the steps.)

|x/3 – 1| < 2
(1/3) * |x – 3| < 2
|x – 3| < 6

The distance of x from 3 is less than 6, so -3 < x < 9. Now we know that every value that x can take will lie within this range.

The question now becomes: what must be true for each of these values of x? Let’s assess each of our answer options with this question:

(A) x > 0
Will each of the values of x be positive? No – x could be a negative number greater than -3, such as -2.

(B) x < 8
Will each of the values of x be less than 8? No – x could be a number between 8 and 9, such as 8.5

(C) x > -4
Will each of the values of x be more than -4? Yes! x will take values ranging from -3 to 9, and each of the values within that range will be greater than -4. So this must be true.

(D) 0 < x < 3
Will each of these values be between 0 and 3. No – since x can take any of the values between -3 and 9, not all of these will be just between 0 and 3.

Therefore, the answer is C (we don’t even need to evaluate answer choice E since C is true).

Again, we have two linked inequalities, but here the relation between them will be a bit different. One of the inequalities is −1 < x < 5 and the other will be the correct answer choice.

We are given that -1 < x < 5, so x lies between -1 and 5. We need an answer choice that “could be true”. This means only some of the values between -1 and 5 should satisfy the condition set by the correct answer choice – all of the values need not satisfy. Let’s evaluate our answer options:

(A) 2x > 10
x > 5
No values between -1 and 5 will be greater than 5, so this cannot be true.

(B) x > 17/3
x > 5.67
No values between -1 and 5 will be greater than 5.67, so this cannot be true.

(C) x^2 > 27
x^2 – 27 > 0
x > 3*√(3) or x < -3*√(3)
√(3) is about 1.73 so 3*1.73 = 5.19. No value of x will be greater than 5.19. Also, -3*1.73 will be -5.19 and no value of x will be less than that. So this cannot be true.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

In today’s post, we will discuss some special formats when we assume variables on the GMAT. These will allow us to minimize the amount of manipulations and calculations that are required to solve certain Quant problems.

Here are some examples:

An even number: 2a
Logic: It must be a multiple of 2.

An odd number: (2a + 1) or (2a – 1)
Logic: It will not be a multiple of 2. Instead, it will be 1 more (or we can say 1 less) than a multiple of 2.

Two consecutive integers: 2a, (2a + 1) or (2a – 1), 2a
Logic: One number will be even and the other will be the next odd number (or the other way around).

Four consecutive odd numbers: (2a – 3), (2a – 1), (2a + 1), (2a + 3)
In this case, the sum of the numbers comes out to be a clean 8a. This can be very useful in many cases.

Five consecutive even numbers: (2a – 4), (2a – 2), 2a, (2a + 2), (2a + 4)
In this case, the sum of the numbers comes out to be a clean 10a. This can also be very useful in many cases.

A prime number: (6a+1) / (6a – 1)
Every prime number greater than 3 is of the form (6a + 1) or (6a – 1). Note, however, that every number of this form is not prime.

Three consecutive numbers:
If we know one number is even and the other two are odd, we will have: (2a – 1), 2a, (2a + 1).
Logic: They add up to give 6a.
In a more generic case, we will have: 3a, (3a+1), (3a+2).
This gives us some important information. It tells us that one of the numbers will definitely be a multiple of 3 and the other two numbers will not be. Note that the numbers can be in a different order such as (3a + 1), (3a + 2) and (3a + 3). (3a + 3) can be written as 3b, so the three numbers will still have the same properties.

Basically, try to pick numbers in a way that will make it easy for you to manage them. Remember, three numbers do not need to be a, b and c – there could be, and in fact often are, several other hints which will give you the relations among the numbers.

Now, let’s see how picking the right format of these numbers can be helpful using a 700-level GMAT question:

The sum of four consecutive odd numbers is equal to the sum of 3 consecutive even numbers. Given that the middle term of the even numbers is greater than 101 and lesser than 200, how many such sequences can be formed?

Now let’s have the three consecutive even numbers be the following, where “b” is any integer: (2b – 2), 2b, (2b + 2)

The sum of these numbers is: (2b – 2) + 2b + (2b + 2) = 6b

Note here that instead of 2a, we used 2b. There is no reason that the even numbers would be right next to the odd numbers, hence we used different variables so that we don’t establish relations that don’t exist between these seven numbers.

We are given that the sum 8a is equal to the sum 6b.

8a = 6b, or a/b = 3/4, where a and b can be any integers. So “a” has to be a multiple of 3 and “b” has to be a multiple of 4.

With this in mind, possible solutions for a and b are:

a = 3, b = 4;
a = 6, b = 8;
a = 9, b = 12
etc.

We are also given that the middle term of the even numbers is greater than 101 and less than 200.

So 101 < 2b < 200, i.e. 50.5 < b < 100.

B must be an integer, hence, 51 ≤ b ≤ 99.

Also, b has to be a multiple of 4, so the values that b can take are 52, 56, 60, 64 … 96

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Test-takers often ask for tips and short cuts to cut down the amount of work necessary to solve a GMAT problem. As such, the Testmaker might want to award the test-taker who pays attention to detail and puts in the required effort.

Today, we will look at an example of this concept – if it seems to be too easy, it is a trap!

In the figure given above, the area of the equilateral triangle is 48. If the other three figures are squares, what is the perimeter, approximately, of the nine-sided shape they form?

(A) 8√(2) (B) 24√(3) (C) 72√(2) (D) 144√(2) (E) 384

The first thing I notice about this question is that we have an equilateral triangle. So I am thinking, the area = s^2 * √(3)/4 and/or the altitude = s*√(3)/2.

The irrational number in play is √(3). There is only one answer choice with √(3) in it, so will this be the answer?

Now, it actually makes me uncomfortable that there is only one option with √(3). At first glance, it seems that the answer has been served to us on a silver plate. But the question format doesn’t seem very easy – it links two geometrical figures together. So I doubt very much that the correct answer would be that obvious.

The next step will be to think a bit harder:

The area of the triangle has √(3) in it, so the side would be a further square root of √(3). This means the actual irrational number would be the fourth root of 3, but we don’t have any answer choice that has the fourth root of 3 in it.

Let’s go deeper now and actually solve the question.

The area of the equilateral triangle = Side^2 * (√(3)/4) = 48

Side^2 = 48*4/√(3)
Side^2 = 4*4*4*3/√(3)
Side = 8*FourthRoot(3)

Now note that the side of the equilateral triangle is the same length as the sides of the squares, too. Hence, all sides of the three squares will be of length 8*FourthRoot(3).

All nine sides of the figure are the sides of squares. Hence:

The perimeter of the nine sided figure = 9*8*FourthRoot(3)
The perimeter of the nine sided figure =72*FourthRoot(3)

Now look at the answer choices. We have an option that is 72√(2). The other answer choices are either much smaller or much greater than that.

Think about it – the fourth root of 3 = √(√(3)) = √(1.732), which is actually very similar to √(2). Number properties will help you figure this out. Squares of smaller numbers (that are still greater than 1) are only a bit larger than the numbers themselves. For example:

(1.1)^2 = 1.21
(1.2)^2 = 1.44
(1.3)^2 = 1.69
(1.414)^2 = 2

Since 1.732 is close to 1.69, the √(1.732) will be close to the √(1.69), i.e. 1.3. Also, √(2) = 1.414. The two values are quite close, therefore, the perimeter is approximately 72√(2). This is the reason the question specifically requests the “approximate” perimeter.

We hope you see how the Testmaker could sneak in a tempting answer choice – beware the “easiest” option!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

We will continue our holistic approach to absolute values and add more complications to these types of questions. This article should set you up for any question of this kind. Note that this is a 750+ level concept, so if you are targeting a lower score, it may not be necessary for you to know.

Using the same logic as we did in the previous two posts, we will word the inequality like this: the distance from 6 should be more than three times the distance from -2.

At x = -2, the distance from 6 is 8 and the distance from -2 is 0. This means the distance from 6 is more than three times the distance from -2.

At x = -1, the distance from 6 is 7 and the distance from -2 is 1. Three times the distance from -2 is 3. This means the distance from 6 is more than three times the distance from -2.

At some point on the right of -1, the distance from 6 will be equal to three times the distance from -2. The distance between -2 and 6 is 8. If we split this 8 into 4 equal parts to get to x = 0, the distance from 6 will be equal to three times the distance from -2.

Now for every point to the right of 0, the distance from 6 will be less than three times the distance from -2.

Let’s try to go to the left of -2 instead. Will there be a point to the left of -2 where the distance from 6 will be equal to three times the distance from -2? Say that point is “a” units away from -2. -2 must then be 2a units away from 6 to ensure that 6 is a total of 3a units away from that point.

The distance between -2 and 6 is 8 – this 8 needs to be equal to 2a, so “a” must be 4 units.

The point where the distance from 6 will be equal to three times the distance from -2 will be 4 units to the left of -2, i.e. at -6. So at points to the right of -6 (but left of 0), the distance from 6 will be more than three times the distance from -2.

Note that for all values to the left of -6, the distance from 6 will be less than three times the distance from -2.

Hence, our x will lie in the range from -6 to 0.

-6 < x < 0

With these parameters, we will have 5 integer solutions: -5, -4, -3, -2 and -1. Hence, our answer is C.

Let’s look at a second question:

For how many integer values of x, is |x – 8| + |5 – x| > |x + 7|?

(A) 1(B) 3(C) 5(D) 7(E) Infinite

Now the true value of this method is visible, as we have three or more terms. The arduous algebra involved in this given inequality makes our logical approach much more attractive.

First note that we have the term |5 – x|. This is the same as |x – 5| because |x| = |-x|.

We will word the inequality like this: the distance from 5 + the distance from 8 should be greater than the distance from -7.

Let’s find the point where the sum of the distance from 5 and the distance from 8 is equal to distance from -7. Say that point is “a” units to the left of 5.

a + a + 3 = 12 – a
a = 3

So the point is 3 units to the left of 5, which means it is at 2. For all points to the left of 2, the sum of the distance from 5 and the distance from 8 will be greater than the distance from -7.

How about the points that are to the right of 8? Say there is a point “b” units away from 8 where the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.

3 + b + b = 15 + b
b = 12

So if we go 12 units to the right of 8, i.e. at x = 20, the sum of the distance from 5 and the distance from 8 is equal to the distance from -7.

For all points to the right of 20, the sum of the distance from 5 and the distance from 8 is greater than the distance from -7, so there will be infinite points for which the sum of the distance from 5 and the distance from 8 is greater than the distance from -7. Therefore, our answer is E.

Using this concept, try to answer the following question on your own: For how many integer values of x, is |x – 6| – |3x + 6| > 0?

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such asthis blog!

Last week, we looked at some absolute value questions involving inequalities. Today, we’ll continue this discussion by adding some more complications to our questions. Consider the question: What is the minimum value of the expression |x – 3| + |x + 1| + |x|? Technically, |x – 3| + |x + 1| + |x| is the sum of “the distance of x from 3,” “the distance of x from -1” and “the distance of x from 0.” To make solving such questions simpler, we’ll often use a parallel situation:

Imagine that there are 3 friends with houses at points -1, 0 and 3 in a straight line. They decide to meet at the point x.

|x – 3| will be the distance covered by the friend at 3 to reach x.

|x + 1| will be the distance covered by the friend at -1 to reach x.

|x| will be the distance covered by the friend at 0 to reach x.

So, the total distance the friends will cover to meet at x will be |x – 3| + |x + 1| + |x|.

Now we can choose to minimize this total distance, bring it to some particular value or make it more or less than some particular value.

If we want to minimize the total distance, we just make the friends meet at the second guy’s house, i.e. at the point 0. The friend at 3 and the friend at -1 need to travel 4 units total to meet anyway, so there’s no point in making the guy at 0 travel any distance at all. So the minimum total distance would be 4, which would then be the minimum value of |x – 3| + |x + 1| + |x|. This minimum value is given by the expression at x = 0.

With this in mind, when we move to the right or to the left of x = 0, the total distance will increase and, hence, the value of the expression |x – 3| + |x + 1| + |x| will also increase.

In our parallel situation of friends and houses, we now have 4 friends with houses at points -1, 0, 5 and 7.

The friends at -1 and 7 are 8 units apart, so they will need to cover at least this total distance together to meet. It doesn’t matter where they meet between -1 and 7 (inclusive), they will need to cover exactly 8 units.

The friends at 0 and 5 will need to travel a minimum distance of 5 to meet. They can meet anywhere between 0 and 5 (inclusive) and the distance they will cover will still be 5.

So, all four friends can meet anywhere between 0 and 5 (inclusive) and the total distance covered will be 8 + 5 = 13. This would be the minimum total distance, and hence, the minimum value of the expression |x – 5| + |x + 1| + |x| + |x – 7|.

When we move to the left of 0 or to the right of 5, the total distance covered will be more than 13. At any point between -1 and 7, the total distance covered by the friends at -1 and 7 will be only 8. When we move 1 unit to the left of 0 and reach -1, the total distance covered by the friends at 0 and 5 will be 1 + 6 = 7. So to meet at -1, the total distance traveled by all friends together will be 8 + 7 = 15.

Similarly, when we move 1 unit to the right of 5 and reach 6, the total distance covered by the four friends will be again 8 + 7 = 15. So at points x = -1 and x = 6, the value of the expression will be 15. Between these two points (excluding the points themselves), the value of the expression will be less than 15.

So now we know -1 < x < 6. With these parameters, x can take 6 integer values: 0, 1, 2, 3, 4, 5. Therefore, the answer is D.

Note that when we had 3 points on the number line, the minimum total distance was found at the second point. Now when we have 4 points on the number line, the minimum total distance has been found to be in the range between second and third points.

Let’s look at another question:

For how many integer values of x, is |2x – 5| + |x + 1| + |x| < 10?

(A) 1(B) 2(C) 4(D) 5(E) Infinite

|2x – 5| + |x + 1| + |x| < 10

2*|x – 5/2| + |x + 1| + |x| < 10

In this sum, now the distance from 5/2 is added twice.

In our parallel situation, this is equivalent to two friends living at 5/2, one living at 0 and one living at -1. Now note that the expression may not take the minimum value of x = 0 because there are 2 people who will need to travel from 5/2.

We have four friends in all, so we can expect to get a range in which we will get the minimum value of the expression. The second and third people are at 0 and 5/2, respectively.

The total distance at x = 0 will be 1 + 2*(5/2) = 6. The total distance at x = 5/2 will be 7/2 + 5/2 = 6.

So if we move to the left of 0 or to the right of 5/2, the total distance will increase. If we move 1 unit to the right of 5/2 and reach 7/2, the total distance covered by the four friends will be 9/2 + 7/2 + 2 = 10. If we move 1 unit to the left of 0 and reach -1, the total distance covered by the four friends will be 0 + 1 + 2*(7/2) = 8. Now all four friends are at -1. To cover a distance of another 2, they should move another 0.5 units to the left of -1 to reach -1.5.

Now the total distance covered by the four friends will be 0.5 + 1.5 + 2*4 = 10, so the total distance when x lies between the points -1.5 and 3.5 (excluding the points themselves) will be less than 10.

Now we know -1.5 < x < 3.5. With these parameters, x can take 5 integer values: -1, 0, 1, 2 and 3. Therefore, the answer is D.

Now use these concepts to solve the following question: For how many integer values of x, is |3x – 3| + |2x + 8| < 15?

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

In an earlier post, I wrote about the GMAT’s tendency to ask questions regarding the number properties of two two-digit numbers whose tens and units digits have been reversed.

The biggest takeaways from that post were:

Anytime we add two two-digit numbers whose tens and units digits have been reversed, we will get a multiple of 11.

Anytime we take the difference of two two-digit numbers whose tens and units digits have been reversed, we will get a multiple of 9.

For the hardest GMAT questions, we’re typically mixing and matching different types of number properties and strategies, so it can be instructive to see how the above axioms might be incorporated into such problems.

Take this challenging Data Sufficiency question, for instance:

When the digits of two-digit, positive integer M are reversed, the result is the two-digit, positive integer N. If M > N, what is the value of M?

(1) The integer (M –N) has 12 unique factors.

(2) The integer (M –N) is a multiple of 9.

The average test-taker looks at Statement 1, sees that it will be very difficult to simply pick numbers that satisfy this condition, and concludes that this can’t possibly be enough information. Well, the average test-taker also scores in the mid-500’s, so that’s not how we want to think.

First, let’s concede that Statement 1 is a challenging one to evaluate and look at Statement 2 first. Notice that Statement 2 tells us something we already know – as we saw above, anytime you have two two-digit numbers whose tens and units digits are reversed, the difference will be a multiple of 9. If Statement 2 is useless, we can immediately prune our decision tree of possible correct answers. Either Statement 1 alone is sufficient, or the statements together are not sufficient, as Statement 2 will contribute nothing. So right off the bat, the only possible correct answers are A and E.

If we had to guess, and we recognize that the average test-taker would likely conclude that Statement 1 couldn’t be sufficient, we’d want to go in the opposite direction – this question is significantly more difficult (and interesting) if it turns out that Statement 1 gives us considerably more information than it initially seems.

In order to evaluate Statement 1, it’s helpful to understand the following shortcut for how to determine the total number of factors for a given number. Say, for example, that we wished to determine how many factors 1000 has. We could, if we were sufficiently masochistic, simply list them out (1 and 1000, 2 and 500, etc.). But you can see that this process would be very difficult and time-consuming.

Alternatively, we could do the following. First, take the prime factorization of 1000. 1000 = 10^3, so the prime factorization is 2^3 * 5^3. Next, we take the exponent of each prime base and add one to it. Last, we multiply the results. (3+1)*(3+1) = 16, so 1000 has 16 total factors. More abstractly, if your number is x^a * y^b, where x and y are prime numbers, you can find the total number of factors by multiplying (a+1)(b+1).

Now let’s apply this process to Statement 1. Imagine that the difference of M and N comes out to some two-digit number that can be expressed as x^a * y^b. If we have a total of 12 factors, then we know that (a+1)(b+1) = 12. So, for example, it would work if a = 3 and b = 2, as a + 1 = 4 and b + 1 = 3, and 4*3 =12. But it would also work if, say, a = 5 and b = 1, as a + 1 = 6 and b + 1 = 2, and 6*2 = 12. So, let’s list out some numbers that have 12 factors:

2^3 * 3^2 (3+1)(2+1) = 12

2^5 * 3^1 (5+1)(1+1) = 12

2^2 * 3^3 (2+1)(3+1) = 12

Now remember that M – N, by definition, is a multiple of 9, which will have at least 3^2 in its prime factorization. So the second option is no longer a candidate, as its prime factorization contains only one 3. Also recall that we’re talking about the difference of two two-digit numbers. 2^2 * 3^3 is 4*27 or 108. But the difference between two positive two-digit numbers can’t possibly be a three-digit number! So the third option is also out.

The only possibility is the first option. If we know that the difference of the two numbers is 2^3 * 3^2, or 8*9 = 72, then only 91 and 19 will work. So Statement 1 alone is sufficient to answer this question, and the answer is A.

Algebraically, if M = 10x + y, then N = 10y + x.

M – N = (10x + y) – (10y + x) = 9x – 9y = 9(x – y).

If 9(x – y) = 72, then x – y = 8. If the difference between the tens and units digits is 8, the numbers must be 91 and 19.

Takeaway: the hardest GMAT questions will require a balance of strategy and knowledge. In this case, we want to remember the following:

Anytime we take the difference of two two-digit numbers whose tens and units digits have been reversed, we will get a multiple of 9.

If one statement is easier to evaluate than the other, tackle the easier one first. If it’s the case that one statement gives you absolutely nothing, and the other is complex, there is a general tendency for the complex statement alone to be sufficient.

For the number x^a * y^b, where x and y are prime numbers, you can find the total number of factors by multiplying (a+1)(b+1).

Instead of looking at how to solve equations, like we did in our previous post, we will look at how to solve inequalities using the same concept.

A quick review:

|x| = The distance of x from 0 on the number line. For example, if |x| = 4, x is 4 away from 0. So x can be 4 or -4.

|x – 1| = The distance of x from 1 on the number line. For example, if |x – 1| = 4, x is 4 away from 1. so x can be 5 or -3.

|x| + |x – 1| = The sum of distance of x from 0 and distance of x from 1 on the number line. for example, if x = 5, the distance of x from 0 is 5 and the distance of x from 1 is 4. The sum of the distances is 5 + 4 = 9. So |x| + |x – 1| = 5 + 4 = 9.

Let’s move ahead now and see how we can use these concepts to solve inequalities:

For how many integer values of x, is |x – 3| + |x + 1| + |x| < 10?

(A) 0 (B) 2 (C) 4 (D) 6 (E) Infinite

In the previous post, we saw the a similar question, except it involved an equation rather than an inequality. For that problem, we found that the two points where the total distance is equal to 10 are -2.667 and 4:

What will be the total distance at any value of x between these two points?

Say, x = 0
|x – 3| + |x + 1| + |x|
= 3 + 1 + 0
= 4

Say, x = 3
|x – 3| + |x + 1| + |x|
0 + 4 + 3
= 7

In both cases, we see that the total distance covered is less than 10. Note that the minimum distance covered will be 4 at x = 0 (discussed in the previous post) so by moving to the right of 0 or to the left of 0 on the number line, we get to the points where the distance increases to 10. So for every point in between, the total distance will be less than 10 (the entire red region).

Hence, at integer points x = -2, -1, 0, 1, 2 and 3 (which are all between -2.667 and 4), the total distance will be less than 10. The total distance will be less than 10 for all non-integer points lying between -2.667 and 4 too, but the question only asks for the integer values, so that is all we need to focus on. (Of course, there are infinite non-integer points between any two distinct points on the number line.) Hence, the answer will be 6 points, or D.

Along the same lines, consider a slight variation of this question:

For how many integer values of x, is |x – 3| + |x + 1| + |x| > 10?

(A) 0 (B) 2 (C) 4 (D) 6 (E) Infinite

What will the answer be here? We hope you immediately jumped to answer choice E – for every integer value of x to the right of 4 or to the left of -2.667, the total distance will be more than 10 (the blue regions). So there will be infinite such integer points (all integers greater than 4 or less than -2.667). Thus, the answer is E.

We hope this logic is clear. We will look at some other variations of this concept next week!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Today we will discuss a problem we sometimes face while attempting to solve Data Sufficiency questions for which the answer is actually E (when both statements together are not sufficient to answer the question). Ideally, we would like to find two possible answers to the question asked so that we know that the data of both statements is not sufficient to give us a unique answer. But what happens when it is not very intuitive or easy to get these two distinct cases?

Let’s try to answer these questions in today’s post using using one of our own Data Sufficiency questions.

A certain car rental agency rented 25 vehicles yesterday, each of which was either a compact car or a luxury car. How many compact cars did the agency rent yesterday?

(1) The daily rental rate for a luxury car was $15 higher than the rate for a compact car.(2) The total rental rates for luxury cars was $105 higher than the total rental rates for compact cars yesterday

We know from the question stem that the total number of cars rented is 25. Now we must find how many compact cars were rented.

There are four variables to consider here:

Number of compact cars rented (this is what we need to find)

Number of luxury cars rented

Daily rental rate of compact cars

Daily rental rate of luxury cars

Let’s examine the information given to us by the statements:

Statement 1: The daily rental rate for a luxury car was $15 higher than the rate for a compact car.

This statement gives us the difference in the daily rental rates of a luxury car vs. a compact car. Other than that, we still only know that a total of 25 cars were rented. We have no data points to calculate the number of compact cars rented, thus, this statement alone is not sufficient. Let’s look at Statement 2:

Statement 2: The total rental rates for luxury cars was $105 higher than the total rental rates for compact cars yesterday.

This statement gives us the difference in the total rental rates of luxury cars vs. compact cars (we do not know the daily rental rates). Again, we have no data points to calculate the number of compact cars rented, thus, this statement alone is also not sufficient.

Now, let’s try to tackle both statements together:

The daily rate for luxury cars is $15 higher than it is for compact cars, and the total rental rates for luxury cars is $105 higher than it is for compact cars. What constitutes this $105? It is the higher rental cost of each luxury car (the extra $15) plus adjustments for the rent of extra/fewer luxury cars hired. That is, if n compact cars were rented and n luxury cars were rented, the extra total rental will be 15n. But if more luxury cars were rented, 105 would account for the $15 higher rent of each luxury car and also for the rent of the extra luxury cars.

Event with this information, we still should not be able to find the number of compact cars rented. Let’s find 2 cases to ensure that answer to this question is indeed E – the first one is quite easy.

We start with what we know:

The total extra money collected by renting luxury cars is $105.

105/15 = 7

Say out of 25 cars, 7 are luxury cars and 18 are compact cars. If the rent of compact cars is $0 (theoretically), the rent of luxury cars is $15 and the extra rent charged will be $105 (7*15 = 105) – this is a valid case.

Now how do we get the second case? Think about it before you read on – it will help you realize why the second case is more of a challenge.

Let’s make a slight change to our current numbers to see if they still fit:

Say out of 25 cars, 8 are luxury cars and 17 are compact cars. If the rent of compact cars is $0 and the rent of luxury cars is $15, the extra rent charged should be $15*8 = $120, but notice, 9 morecompact cars were rented than luxury cars. In reality, the extra total rent collected is $105 – the $15 reduction is because of the 9 additional compact cars. Hence, the daily rental rate of each compact car would be $15/9 = $5/3.

This would mean that the daily rental rate of each luxury car is $5/3 + $15 = $50/3

The total rental cost of luxury cars in this case would be 8 * $50/3 = $400/3

The total rental cost of compact cars in this case would be 17 * $5/3 = $85/3

The difference between the two total rental costs is $400/3 – $85/3 = 315/3 = $105

Everything checks out, so we know that there is no unique answer to this question – for any number of compact cars you use, you will come up with the same answer. Thus, Statements 1 and 2 together are not sufficient.

The strategy we used to find this second case to test is that we tweaked the numbers we were given a little and then looked for a solution. Another strategy is to try plugging in some easy numbers. For example:

Instead of using such difficult numbers, we could have tried an easier split of the cars. Say out of 25 cars, 10 are luxury and 15 are compact. If the rent of compact cars is $0 and the rent of luxury cars is $15, the extra rent charged should be 10*$15 = $150 extra, but it is actually only $105 extra, a difference of $45, due to the 5 additional compact cars. The daily rental rent of 5 extra compact cars would be $45/5 = $9. Using these numbers in the calculations above, you will see that the difference between the rental costs is, again, $105. This is a valid case, too.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

The non-election trending story of the day is the announcement of the forthcoming Nintendo Switch gaming system, a system that promises to help you take the utmost advantage of your leisure time…but that may help you maximize the value of your GMAT experience, too.

How?

The main feature of the Switch (and the driving factor behind its name) is its flexibility. It can be an in-home gaming system attached to a fixed TV set, but then immediately Switch to a hand-held portable system that allows you to continue your game on the go. Nintendo’s business plan is primarily based on offering flexibility…and on the GMAT, your plan should be to prove to business schools that you can offer the same.

The GMAT, of course, tests algebra skills and critical thinking skills and grammar skills, but beneath the surface it also has a preference for testing flexibility. Many problems will punish those with pure tunnel vision, but reward those who can identify that their first course of action isn’t working and who can then Switch to another plan. This often manifests itself in:

Math problems that seem to require algebra…but halfway through beg to be back-solved using answer choices.

Sentence Correction problems that seem to ask you to make a decision about one major difference…but for which the natural choices leave you with clearer-cut errors elsewhere.

Critical Reasoning answer choices that seem out of scope at first, but reward those who read farther and then see their relevance.

Data Sufficiency problems for which you’ve made a clear, confident decision on one statement…but then the other statement shows you something you hadn’t considered before and forces you to reconsider.

The overall concept that if you’re a one-trick pony – you’re a master of plugging in answer choices, for example – you’ll find questions that just won’t reward that strategy and will force you to do something else.

Flexibility matters on the GMAT! As an example, consider the following Data Sufficiency question:

Is x/y > 3?

1) 3x > 9y2) y > 3y

If you’re like many, you’ll confidently address the algebra in Statement 1, divide both sides by 3 to get x > 3y, and then see that if you divide both sides by y, you can make it look exactly like the question stem: x/y > 3. And you may very well say, “Statement 1 is sufficient!” and confidently move on to Statement 2.

But when you look at Statement 2 – either conceptually or algebraically – something should stand out. For one, there’s no way that it’s sufficient because it doesn’t help you determine anything about x. And secondly, it brings up the point that “y is negative” (algebraically you’d subtract y from both sides to get 0 > 2y, then divide by 2 to get 0 > y). And here’s where, if it hasn’t already, your mind should Switch to “positive/negative number properties” mode. If you weren’t thinking about positive vs. negative properties when you considered Statement 1, this one gives you a chance to Switch your thinking and reconsider – what if y were negative? Algebraically, you’d then have to flip the sign when you divide both sides by y:

3x > 9y : Divide both sides by 3

x > 3y : Now divide both sides by y, but remember that if y is positive you keep the sign (x/y > 3), and if y is negative you flip the sign (x/y < 3).

With this in mind, Statement 1 doesn’t really tell you anything. x/y can be greater than 3 or less than 3, so all Statement 1 does is eliminate that x/y could be exactly 3. Now you have the evidence to Switch your answer. If you initially thought Statement 1 was sufficient, Statement 2 has given you a chance to reassess (thereby demonstrating flexibility in thinking) and realize that it’s not, until you know whether y is negative or positive.

Statement 2 supplies that missing piece, and the answer is thus C. But more important is the lesson – because the GMAT so values mental flexibility, it will often provide you with clues that can help you change your mind if you’re paying attention. So on the GMAT, take a lesson from Nintendo Switch: flexibility is an incredibly marketable skill, so look for clues and opportunities to Switch your line of thinking and save yourself from trap answers.

The moment we see an equation involving the variable x, we have a habit of jumping right into attempting to solve it. But what happens when we are not able to solve it? Let’s say, for example, we have an equation such as x^2 + 1 = 0. How would we solve for x here? We can’t because x has no real value. Note that x^2 is non-negative – it would be either 0 or positive. 1, we know, is positive. So together, a positive number and a non-negative number cannot add up to 0.

In this example, it relatively easy to see that the equation has no real solution. In others, it may not be so obvious, so we will need to use other strategies.

We know how to solve third degree equations. The first solution is found by trial and error – we try simple values such as -2, -1, 0, 1, 2, etc. and are usually able to find the first solution. Then the equation of third degree is split into two factors, including a quadratic. We know how to solve a quadratic, and that is how we get all three solutions, if it has any.

But what if we are unable to find the first solution to a third degree equation by trial and error? Then we should force ourselves to wonder if we even need to solve the equation at all. Let’s take a look at a sample question to better understand this idea:

Is x < 0?(1) x^3 + x^2 + x + 2 = 0(2) x^2 – x – 2 < 0

In this problem, x can be any real number – we have no constraints on it. Now, is x negative?

Statement 1: x^3 + x^2 + x + 2 = 0

If we try to solve this equation as we are used to doing, look at what happens:

If you plug in x = 2, you get 16 = 0
If you plug in x = 1, you get 5 = 0
If you plug in x = 0, you get 2 = 0
If you plug in x = -1, you get 1 = 0
If you plug in x = -2, you get -4 = 0

We did not find any root for the equation. What should we do now? Note that when x goes from -1 to -2, the value on the left hand side changes from 1 to -4, i.e. from a positive to a negative. So, in between -1 and -2 there will be some value of x for which the left hand side will become 0. That value of x will not be an integer, but some decimal value such as -1.3 or -1.4, etc.

Even after we find the first root, making the quadratic will be very tricky and then solving it will be another uphill task. So we should ask ourselves whether we even need to solve this equation.

Think about it – can x be positive? If x is indeed positive, x^3, x^2 and x all will be positive. Then, if we add four positive numbers (x^3, x^2, x and 2) we will get a positive sum – we cannot get 0. Obviously x cannot be 0 since that will give us 2 = 0.

This means the value of x must be negative, but what it is exactly doesn’t matter. We know that x has to be negative, and that is sufficient to answer the question.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

What determines whether or not a question can be considered a GMAT question? We know that GMAT questions that are based on seemingly basic concepts can be camouflaged such that they may “appear” to be very hard. Is it true that a question requiring a lot of intricate calculations will not be tested in GMAT? Yes, however it is certainly possible that a question may “appear” to involve a lot of calculations, but can actually be solved without any!

In the same way, it is possible that a question may appear to be testing very obscure concepts, while it is really solvable by using only basic ones.

This happens with one of our own practice questions – we have often heard students exclaim that this problem isn’t relevant to the GMAT since it “tests an obscure number property”. It is a question that troubles many people, so we decided to tackle it in today’s post.

We can easily solve this problem with just some algebraic manipulation, without needing to know any obscure properties! Let’s take a look:

† and ¥ represent non-zero digits, and (†¥)² – (¥†)² is a perfect square. What is that perfect square?

(A) 121 (B) 361 (C) 576 (D) 961 (E) 1,089

The symbols † and ¥ are confusing to work with, so the first thing we will do is replace them with the variables A and B.

The question then becomes: A and B represent non-zero digits, and (AB)² – (BA)² is a perfect square. What is that perfect square?

As I mentioned before, we have heard students complain that this question isn’t relevant to the GMAT because it “uses an obscure number property”. Now here’s the thing – most advanced number property questions CAN be solved in a jiffy using some obscure number property such as, “If you multiply a positive integer by its 22nd multiple, the product will be divisible by …” etc. However, those questions are not actually about recalling these so-called “properties” – they are about figuring out the properties using some generic technique, such as pattern recognition.

For this question, the complaint is often that is that the question tests the property, “(x + y)*(x – y) (where x and y are two digit mirror image positive integers) is a multiple of 11 and 9.” It doesn’t! Here is how we should solve this problem, instead:

Given the term (AB)^2, where A and B are digits, how will you square this while keeping the variables A and B?

Let’s convert (AB)^2 to (10A + B)^2, because A is simply the placeholder for the tens digit of the number. If you are not sure about this, consider the following:

We know now that the expression is a multiple of 9 and 11. We would not have known this beforehand. Now we’ll just use the answer choices to figure out the solution. Only 1,089 is a multiple of both 9 and 11, so the answer must be E.

We hope you see that this question is not as hard as it seems. Don’t get bogged down by unknown symbols – just focus on the next logical step at each stage of the problem.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such asthis blog!

If you’re like many – to the dismay of the NFL and the advertising industry – you’re planning to watch another presidential debate this coming Sunday. And just like Trump-Clinton I and Pence-Kaine earlier this week, this debate will provide plenty of opportunities to be annoyed, frustrated, and disappointed…but it will also provide an ever-important lesson about the GMAT.

It’s no surprise that candidate approval ratings are low for the same reason that far too many GMAT scores are lower than candidates would hope. Why?

People don’t directly answer the question.

This is incredibly common in the debates, where the poor moderators are helpless against the talking points and stump speeches of the candidates. The public then suffers because people cannot get direct answers to the questions that matter. This is also very common on the GMAT, where students will invest the time in critical thought and calculation, and then levy an answer that just doesn’t hit the mark. Consider the example:

Donald has $520,000 in campaign money available to spend on advertising for the month of October, and his advisers are telling him that he should spend a minimum of $360,000 in the battleground states of Ohio, Florida, Virginia, and North Carolina. If he plans to spend the minimum amount in battleground states to appease his advisers, plus impress his friends by a big ad spend specific to New York City (and then he will skip advertising in the rest of the country), how much money will he have remaining if he wants 20% of his ad spend to take place in New York City?

(A) $45,000 (B) $52,000 (C) $70,000 (D) $90,000 (E) $104,000

As people begin to calculate, it’s common to try to determine all of the facets of Donald’s ad spend. If he’s spending only the $360,000 in battleground states plus the 20% he’ll spend in New York City, then $360,000 will represent 80% of his total ad spend. If $360,000 = 0.8(Total), then the total will be $450,000. That means that he’ll spend $90,000 in New York City. Which is answer choice D…but that’s not the question!

The question asked for how much of his campaign money would be left over, so the calculation you need to focus on is the $520,000 he started with minus the $450,000 he spent for a total of $70,000, answer choice C. And in a larger context, you can learn a major lesson from Wharton’s most famous alumnus: it’s not enough for your answer to be related to the question. On the GMAT, you must answer the question directly! So make sure that you:

Be careful with Strengthen/Weaken Critical Reasoning problems. A well-written Strengthen problem will likely have a good Weaken answer choice, and vice-versa.

In algebra problems, make sure to identify the proper variable (or combination of variables if they ask, for example “What is 6x – y?”).

With Data Sufficiency problems, pay attention to the exact values being asked for. One of the most common mistakes that people make is saying that a statement is insufficient because they’re looking to fill in all variables, when actually it is sufficient to answer the exact combination that the test asked for.

As you watch the debate this weekend, notice (How could you not?) how absurd it is that the candidates just about never directly answer the question…and then vow to not make the same mistake on your GMAT exam.

It does not surprise anyone when they learn that the properties of circles are tested on the GMAT. Most test-takers will nod and rattle off the relevant equations by rote: Area = Π*radius ^2; Circumference = 2Π* radius; etc. However, many of my students are caught off guard to learn that the equation for a circle on the coordinate plane is our good friend the Pythagorean theorem. Why on earth would an equation for a right triangle describe a circle?

Take a look at the following diagram in which a circle is centered on the origin (0,0) in the coordinate plane:

Designate a random point on the circle (x,y.) If we draw a line from the center of the circle to x,y, that line is a radius of the circle. Call it r. If we drop a line down from (x,y) to the x-axis, we’ll have a right triangle:

Note that the base of the triangle is x, and the height of the triangle is y. So now we have our Pythagorean theorem: x^2 + y^2 = r^2. This is also the equation for a circle centered on the origin on the coordinate plane. [The more general equation for a circle with a center (a,b) is (x-a)^2 + (y-b)^2 = r^2. When a circle is centered on the origin, (a,b) is simply (0,0.)]

This ends up being an immensely useful tool to use on the GMAT. Take the following question, for example:

A certain circle in the xy-plane has its center at the origin. If P is a point on the circle, what is the sum of the squares of the coordinates of P?

(1) The radius of the circle is 4(2) The sum of the coordinates of P is 0

So let’s draw this, designating P as (x,y):

Now we draw our trust right triangle by dropping a line down from P to the x-axis, which will give us this:

We’re looking for x^2 + y^2. Hopefully, at this point, you notice what the question is going for – because we have a right triangle, x^2 + y^2 = r^2, meaning that all we need is the radius!

Statement 1 is pretty straightforward – if r = 4, we can insert this into our equation of x^2 + y^2 = r^2 to get x^2 + y^2 = 4^2. So x^2 + y^2 = 16. Clearly, this is sufficient.

Now look at Statement 2. If the sum of x and y is 0, we can say x = 1 and y = -1 or x = 2 and y = -2 or x = 100 and y = -100, etc. Each of these will yield a different value for x^2 + y^2, so this statement alone is clearly not sufficient. Our answer is A.

Takeaway: any shape can appear on the coordinate plane. If the shape in question is a circle, remember to use the Pythagorean theorem as your equation for the circle, and what would have been a challenging question becomes a tasty piece of baklava. (We are talking about principles elucidated by the ancient Greeks, after all.)

Today, we will give you a GMAT challenge question. The challenge of reviewing this question is not that the question is hard to understand – it is that you will need to solve this official question within a minute using minimum calculations.

Let’s take a look at the question stem:

Date of Transaction

Type of Transaction

June 11

Withdrawal of $350

June 16

Withdrawal of $500

June 21

Deposit of x dollars

For a certain savings account, the table shows the three transactions for the month of June. The daily balance for the account was recorded at the end of each of the 30 days in June. If the daily balance was $1,000 on June 1 and if the average (arithmetic mean) of the daily balances for June was $1,000, what was the amount of the deposit on June 21?

(A) $1,000(B) $1,150(C) $1,200(D) $1,450(E) $1,600

Think about how you might answer this question:

The average of daily balances = (Balance at the end of June 1 + Balance at the end of June 2 + … + Balance at the end of June 30) / 30 = 1000

Now we have been given the only three transactions that took place:

A withdrawal of $350 on June 11 – so on June 11, the account balance goes down to $650.

A withdrawal of $500 on June 16 – so on June 16, the account balance goes down to $150.

A deposit of $x on June 21 – So on June 21, the account balance goes up to 150 + x.

Now we can plug in these numbers to say the average of daily balances = [1000 + 1000 + …(for 10 days, from June 1 to June 10) + 650 + 650 + … (for 5 days, from June 11 to June 15) + 150 + … (for 5 days, from June 16 to June 20) + (150 + x) + (150 + x) + … (for 10 days, from June 21 to June 30)] / 30 = 1000

Note that we are talking about the average of certain data values. Also, we know the deviations from those data values:

The amount from June 11 to June 30 is 350 less.

The amount from June 16 to June 30 is another 500 less.

The amount from June 21 to June 30 is x in excess.

Through the deviation method, we can see the shortfall = the excess:

350 * 20 + 500 * 15 = x * 10
x = 1,450 (D)

This simplifies our calculation dramatically! Though saving only one minute on a question like this may not seem like a very big deal, saving a minute on every question by using a more efficient method could be the difference between a good Quant score and a great Quant score!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

In some Quant questions, we are given big nasty numbers in the answer choices and little else in the question stem. Often in such cases, the starting point is difficult for the test-taker to find, so today, we will discuss how to handle such questions.

The first and only rule with these types of problems is that familiarity helps. Evaluate the answer choices that make sense to you first.

Let’s look at a few questions to understand how to do that:

Which of the following is NOT prime?

(A) 1,556,551(B) 2,442,113(C) 3,893,257(D) 3,999,991(E) 9,999,991

The first thing that comes to mind when we consider how to find prime numbers should be to “check the number N for divisibility by all prime factors until we get to the √N.” But note that here, we have four numbers that are prime and one number that is not. Also, the numbers are absolutely enormous and, hence, will be very difficult to work with. So, let’s slide down to a number that seems a bit more sane: 3,999,991 (it is very close to a number with lots of 0’s).

3,999,991 = 4,000,000 – 9
= (2000)^2 – 3^2

This is something we recognise! It’s a difference of squares, which can be written as:

= (2000 + 3) * (2000 – 3)
= 2003 * 1997

Hence, we see that 3,999,991 is a product of two factors other than 1 and itself, so it is not a prime number. We have our answer! The answer is D.

Let’s try another problem:

Which of the following is a perfect square?

(A) 649 (B) 961 (C) 1,664 (D) 2,509 (E) 100,000

Here, start by looking at the answer choices. The first one that should stand out is option E, 100,000, since multiples of 10 are always easy to handle. However, we have an odd number of zeroes here, so we know this cannot be a perfect square.

Next, we look at the answer choices that are close to the perfect squares that we intuitively know, such as 30^2 = 900, 40^2 = 1600, 50^2 = 2500. The only possible number whose perfect square could be 961 is 31 – 31^2 will end with a 1 and will be a bit greater than 900 (32^2 will end with a 4, so that cannot be the square root of 961, and the perfect squares of other greater numbers will be much greater than 900).

31^2 = (30 + 1)^2 = 900 + 1 + 2*30*1 = 961

So, we found that 961 is a perfect square and is, hence, the answer!

In case 961 were not a perfect square, we would have tried 1,664 since it is just 64 greater than 1,600. It could be the perfect square of 42, as the perfect square of 42 will end in a 4.

If 1,664 were also not a perfect square (it is not), we would have looked at 2,509. We would have known immediately that 2,509 cannot be a perfect square because it is too close to 2,500. 2,509 ends in a 9, so we may have considered 53 to be its square root, but the difference between consecutive perfect squares increases as we get to greater numbers.

(4^2 is 16 while 5^2 is 25 – the difference between them is 9. The difference between 5^2 and 6^2 will be greater than 9, and so will the difference between the perfect squares of any pair of consecutive integers greater than 6. Hence, the difference between the squares of 50 and 53 certainly cannot be 9.)

Therefore, our answer is B. Let’s try one more question:

When a certain perfect square is increased by 148, the result is another perfect square. What is the value of the original perfect square?

(A) 1,296(B) 1,369(C) 1,681(D) 1,764(E) 2,500

This question is, again, on perfect squares. We can use the same concepts here, too.

30^2 = 900
31^2 = 961 (=(30+1)^2 = 900 + 1 + 2*30)

40^2= 1,600
41^2 = 1,681 (=(40+1)^2 = 1,600 + 1 + 2*40)

50^2 = 2,500
51^2 = 2,601 (=(50+1)^2 = 2,500 + 1 + 2*50)

We know that the difference between consecutive squares increases as we go to greater numbers: going from 30^2 to 31^2 is a difference of 61, while jumping from 40^2 to 41^2 is a difference of 81.

All the answer choices lie in the range from 900 to 2500. In this range, the difference between consecutive squares is between 60 and 100. So, when you add 148 to a perfect square to get another perfect square in this range, we can say that the numbers must be 2 apart, such as 33 and 35 or 42 and 44, etc. Also, the numbers must lie between 30 and 40 because twice 61 is 122 and twice 81 is 162 – 148 lies somewhere in between 122 and 162.

A and B are the only two possible options.

Consider option A – it ends in a 6, so the square root must end in a 6, too. If you add 148, then it will end with a 4 (the perfect square of a number ending in 8 will end in 4). So this answer choice works.

Consider option B – it ends in a 9, so the square root must end in a 3 or a 7. When you add 148, it ends in 7. No perfect square ends in 7, so this option is out. Our answer is, therefore, A.

We hope you see how a close evaluation of the answer choices can help you solve questions of this type. Go get ’em!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Arguably the biggest news story this week was presidential hopeful Gary Johnson’s reply to a foreign policy question. “What is Aleppo?” is what Johnson responded, his mind evidently blanking on the epicenter of Syrian civil war and its resulting refugee crisis. And regardless of your opinion of Johnson’s fitness to be the architect of American foreign policy, there’s one major lesson there for your GMAT aspirations:

In pressure situations, it’s not uncommon for your brain to fail you as you “blank” on a concept you know (or should know). So it’s important to have strategies ready for that moment that very well may come to you. To paraphrase the Morning Joe question to Johnson:

What would you do about “Aleppo?”

Meaning: what would you do if your mind were to go blank on an important GMAT rule or formula?

There are four major strategies that should be in your toolkit for such a situation:

1) Test Small Numbers
You should absolutely know formulas like exponent rules or relationships like that between dividend, divisor, and remainder in division, but sometimes your mind just goes blank. In those cases, remember that math rules are logically-derived, not arbitrarily ordained! Math rules will hold for all possible values, so if you’re unsure, test numbers. For example, if you’re forced to solve something like:

(x^15)(x^9) =

And you’ve blanked on what to do with exponents, try testing small numbers like (2^2)(2^3). Here, that’s (4)(8) = 32, which is 2^5. So if you’re unsure, “Do I add or multiply the exponents?” you should see from the small example that you definitely don’t multiply, and that your hunch that, “Maybe I add?” works in this case, so you can more confidently make that decision.

Similarly, if a problem asked:

When integer y is divided by integer z, the quotient is equal to x. Which of the following represents the remainder in terms of x, y, and z?

(A) x – yz (B) zy – x (C) y – zx (D) zy – x (E) zx – y

Many students memorize equations to organize dividend, divisor, quotient, and remainder, but in the fog of war on test day it can even be difficult to remember which element of the division problem is the dividend (it’s the number you start with) and which is the divisor (it’s the one you divide by). So if your mind has blanked on any part of the equation or on which element is which, just test it with small numbers to remind yourself how the concept works:

11 divided by 4 is 2 with a remainder of 3. How do you get to the remainder? You take the 11 you started with and subtract the 8 that you get from taking the divisor of 4 and multiplying it by the quotient of 2. So the answer is y (what you started with) minus zx (the divisor times the quotient), or answer choice C.

Simply put, if you blank on a rule or concept, you can test small numbers to remind yourself how it works.

2) Use Process of Elimination and Work Backwards From the Answer Choices
One beautiful thing about the GMAT is that, while in “the real world” if you need to know the Pythagorean Theorem and blank on it, you’re out of luck (well, unless you have a Google-enabled Smartphone in your pocket which you almost certainly do…), on the GMAT you have answer choices as assets. So if your own work stalls in progress, you can look to the answer choices to eliminate options you know for sure you wouldn’t get with that math:

What is x^5 + x^6? You know you don’t add or multiply those exponents, so even if you don’t see to factor out the common x^5, you could eliminate answer choices like x^11 and x^30.

Or you can look to the answer choices to see if they help you determine how you’d apply a rule. For example, if a problem forces you to employ the side ratios for a 45-45-90 triangle and you’ve forgotten them, the presence of some square roots of 2 in the answer choices can help you remember. The square root of 2 is greater than 1, and two sides must match, so if someone spots you “the rule includes a square root of 2” the only thing it can really be is the ratio x : x : x(√2)

Gary Johnson should have been so lucky – had the question been posed as, “What would you do about Aleppo, which is either a DJ on the new Drake album; the epicenter of the Syrian crisis; or a new restaurant in the Garment District?” he would get that question right every single time. Answer choices are your friends…when you blank, consult them!

3) Think Logically
Similar to that 45-45-90 “what else could it be?” logic, many times when you blank on a rule, you can work your way to either the rule itself or just to the answer by thinking logically about it. For example, if you end up with math that includes a radical sign in the denominator and can’t quite remember the steps for rationalizing the denominator:

What is 1/(1 – √2)?

(A) √2 (B) 1 – √2 (C) 1 + √2 (D) -1 – √2 (E) √2 – 1

Not all is lost! Sure, algebraically you should multiply the numerator and the denominator by the conjugate (1 + √2) but you can also logically work with this one. The numerator is 1, and the denominator is 1 – the square root of 2. You know that √2 is between 1 and 2, so what do you know about the denominator? It’s negative, and it’s a fraction (or decimal), so once you’ve taken 1 divided by that, your answer must be a negative number to the left of -1 – only answer choice D would work. So, yeah, you blanked on the steps, but you can still employ logic to back into the answer.

4) Write Down Everything You Know
Blanking is particularly troublesome because it’s that moment of panic. You’re trying to retrace your mental steps and the answer is elusive; it’s a moment you’re not in control of at that point. So take control! The more you’re actively working – jotting down other related formulas or facts you know, working on other facets of the diagram or problem and saving that step for last, etc. – the more you’re controlling, or at least actively managing, the situation.

Gary Johnson couldn’t get away with a “Who Wants to Be a Millionaire?” style talk-through-it (“Um, I know it’s not the name of any congressmen; it’s not Zika, it’s not…”) without looking dumb, but no one is going to audit your scratchwork and release it to Huffington Post, so you’re free to jot down half-baked thoughts and trial calculations to your heart’s content. Actively manage the situation, and you can work your way through that dreaded “my mind is blank” moment.

So learn from Gary Johnson. No matter how much you’ve prepared for your GMAT, there’s a chance that your mind will go blank on something you know that you know, but just can’t recall in the moment. But you have options, so heed the wisdom above, and let Trump or Clinton handle the gaffes for the day while you move on confidently to the next question.

When test-takers first learn how to tackle combination and permutation questions, there’s typically a moment of euphoria when the proper approach really clicks.

If, for example, there are 10 people in a class, and you wish to find the number of ways you can form a cabinet consisting of a president, a vice president, and a treasurer, all you need to do is recognize that if you have 10 options for the president, you’ll have 9 left for the vice president, and 8 remaining for the treasurer, and the answer is 10*9*8. Easy, right?

But on the GMAT, as in life, anything that seems too good to be true probably is. An easy question can be tackled with the type of mechanical thinking illustrated above. A harder question will require a more sophisticated approach in which we consider disparate scenarios and perform calculations for each.

Take this question, for example:

Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

A) 84B) 91C) 100D) 105E) 243

It’s natural to see this problem and think, “All I have to do is reason out how many options I have for each digit. So for the hundreds digit, I have 3 options (7, 8, or 9); the tens digit has to be different from the hundreds digit, and it must be non-zero, so I’ll have 8 options here; then the last digit has to be odd, so…”

Here’s where the trouble starts. The number of eligible numbers in the 700’s will not be the same as the number of eligible numbers in the 800’s -if the digits must all be different, then a number in the 700’s can’t end in 7, but a number in the 800’s could. So, we need to break this problem into separate cases:

First Case: Numbers in the 700’s
If we’re dealing with numbers in the 700’s, then we’re calculating how many ways we can select a tens digit and a units digit. 7___ ___.

Let’s start with the units digit. Well, we know that this number needs to be odd. And we know that it must be different from the hundreds and the tens digits. This leaves us the following options, as we’ve already used 7 for the hundreds digit: 1, 3, 5, 9. So there are 4 options remaining for the units digit.

Now the tens digit must be a non-zero number that’s different from the hundreds and units digit. There are 9 non-zero digits. We’re using one of those for the hundreds place and one of those for the units place, leaving us 7 options remaining for the tens digit. If there are 4 ways we can select the units digit and 7 ways we can select the tens digit, there are 4*7 = 28 options in the 700’s.

Second Case: Numbers in the 800’s
Same logic: 8 ___ ___. Again, this number must be odd, but now we have 5 options for the units digit, as every odd number will obviously be different from the hundreds digit, which is even (1, 3, 5, 7, or 9). The tens digit logic is the same – 9 non-zero digits total, but it must be different from the hundreds and the units digit, leaving us 7 options remaining. If there are 5 ways we can select the units digit and 7 ways we can select the tens digit, there are 5*7 = 35 options in the 800’s.

Third Case: Numbers in the 900’s
This calculation will be identical to the 700’s scenario: 9___ ___. For the units digit, we want an odd number that is different from the hundreds digit, giving us (1, 3, 5, 7), or 4 options. We’ll have 7 options again for the tens digit, for the same reasons that we’ll have 7 options for the tens digit in our other cases. If there are 4 ways we can select the units digit and 7 ways we can select the tens digit, then there are 4*7 = 28 options in the 900’s.

To summarize, there are 28 options in the 700’s, 35 options in the 800’s, and 28 options in the 900’s. 28 + 35 + 28 = 91. Therefore, B is the correct answer.

Takeaway: for a simpler permutation question, it’s fine to simply set up your slots and multiply. For a more complicated problem, we’ll need to work case-by-case, bearing in mind that each individual case is, on its own, actually not nearly as hard as it looks, sort of like the GMAT itself.

We have discussed how to use the deviation method to find the arithmetic mean of numbers. It is very useful in cases where the numbers are huge, as it considerably brings down the calculation time.

The same method can be applied to weighted averages, as well. Let’s look at an example very similar to the one we examined when we were working on deviations in the case of arithmetic means:

What is the average of 452, 452, 453, 460, 467, 480, 499, 499, 504?

What would you say the average is here? Perhaps, around 470?

Shortfall:
We have two 452s – 452 is 18 less than 470.
453 is 17 less than 470.
460 is 10 less than 470.
467 is 3 less than 470.

Overall, the numbers less than 470 are (2*18) + 17 + 10 + 3 = 66 less than 470.

Excess:
480 is 10 more than 470.
We have two 499s – 499 is 29 more than 470.
504 is 34 more than 470.

Overall, the numbers more than 470 are 10 + (2*29) + 34 = 102 more than 470.

The shortfall is not balanced by the excess; there is an excess of 102-66 = 36.

So what is the average? If we assume that the average of these 9 numbers is 470, there will be an excess of 36. We need to distribute this excess evenly among all of the numbers, and hence, the average will increase by 36/9 = 4.

Therefore, the required mean is 470 + 4 = 474. (If we had assumed the mean to be 474, the shortfall would have balanced the excess.)

This method is used in exactly the same way when we have a simple average as when we have a weighted average. The reason we are reviewing it is that it can be very handy in weighted average questions involving more than two quantities.

We often deal with questions on weighted averages involving two quantities using the scale method. Let’s see how to use the deviation method for more than 2 quantities on an official GMAT question:

Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x+y+z gallons of a 1.5 percent grade, what is x in terms of y and z?

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

As the northern hemisphere drifts toward autumn, two events have become just about synonymous: Labor Day and Back to School. If you’re spending this Labor Day weekend getting yourself ready to go back to graduate school, you may well labor over GMAT study materials in between barbecues and college football games. And if you do, make sure you heed this wisdom: GMAT test day should not be Labor Day!

What does that mean?

On a timed test like the GMAT, one of the biggest drains on your score can be a combination of undue time and undue energy spent on problems that could be done much simpler. “The long way is the wrong way” as a famous GMAT instructor puts it – those seconds you waste, those extra steps that could lead to error or distraction, they’ll add up over the test and pull your score much lower than you’d like it to be. With that in mind, here are six ways to help you avoid too much labor on test day:

QUANTITATIVE SECTION1) Do the math in your order, only when necessary.
Because the GMAT doesn’t allow a calculator, it heavily rewards candidates who can find efficient ways to avoid the kind of math for which you’d need a calculator. Very frequently this means that the GMAT will tempt you with calculations that you’d ordinarily just plug-and-chug with a calculator, but that can be horribly time-consuming once you start.

For example, a question might require you to take an initial number like 15, then multiply by 51, then divide by 17. On a calculator or in Excel, you’d do exactly that. But on the GMAT, that calculation gets messy. 15*51 = 765 – a calculation that isn’t awful but that will take most people a few steps and maybe 20 seconds. But then you have to do some long division with 17 going into 765. Or do you? If you’re comfortable using factors, multiples, and reducing fractions, you can see those two steps (multiply by 51, divide by 17) as one: multiply by 51/17, and since 51/17 reduces to 3, then you’re really just doing the calculation 15*3, which is easily 45.

The lesson? For one, don’t start doing ugly math until you absolutely know you have to perform that step. Save ugly math for later, because the GMAT is notorious for “rescuing” those who are patient enough to wait for future steps that will simplify the process. And, secondly, get really, really comfortable with factors and divisibility. Quickly recognizing how to break a number into its factors (51 = 3*17; 65 = 5*13; etc.) allows you to streamline calculations and do much of the GMAT math in your head. Getting to that level of comfort may take some labor, but it will save you plenty of workload on test day.

2) Recognize that “Answers Are Assets.”
Another way to avoid or shortcut messy math is to look at the answer choices first. Some problems might look like they involve messy algebra, but can be made much easier by plugging in answer choices and doing the simpler arithmetic. Other times, the answer choices will lead themselves to process of elimination, whether because some choices do not have the proper units digit, or are clearly too small.

Still others will provide you with clues as to how you have to attack the math. For example, if the answer choices are something like: A) 0.0024; B) 0.0246; C) 0.246; D) 2.46; E) 24.6, they’re not really testing you on your ability to arrive at the digits 246, but rather on where the decimal point should go (how many times should that number be multiplied/divided by 10). You can then set your sights on the number of decimal places while not stressing other details of the calculation.

Whatever you do, always scan the answer choices first to see if there are easier ways to do the problem than to simply slog through the math. The answers are assets – they’re there for a reason, and often, they’ll provide you with clues that will help you save valuable time.

3) Question the Question – Know where the game is being played.
Very often, particularly in Data Sufficiency, the GMAT Testmaker will subtly provide a clue as to what’s really being tested. And those who recognize that can very quickly focus on what matters and not get lost in other elements of the problem.

For example, if the question stem includes an inequality with zero (x > 0 or xy < 0), there’s a very high likelihood that you’re being tested on positive/negative number properties. So, when a statement then says something like “1) x^3 = 1331”, you can hold off on trying to take the cube root of 1331 and simply say, “Odd exponent = positive value, so I know that x is positive,” and see if that helps you answer the question without much calculation. Or if the problem asks for the value of 6x – y, you can say to yourself, “I may not be able to solve for x and y individually, but if not, let’s try to isolate exactly that 6x – y term,” and set up your algebra accordingly so that you’re efficiently working toward that specific goal.

Good test-takers tend to see “where the game is being played” by recognizing what the Testmaker is testing. When you can see that a question is about number properties (and not exact values) or a combination of values (and not the individual values themselves) or a comparison of values (again, not the actual values themselves), you can structure your work to directly attack the question and not fall victim to a slog of unnecessary calculations.

VERBAL SECTION4) Focus on keywords in Critical Reasoning conclusions.
The Verbal section simply looks time-consuming because there’s so much to read, so it pays to know where to spend your time and focus. The single most efficient place to spend time (and the most disastrous if you don’t) is in the conclusion of a Strengthen or Weaken question. To your advantage, noticing a crucial detail in a conclusion can tell you exactly “where the game is being played” (Oh, it’s not how much iron, it’s iron PER CALORIE; it’s not that Company X needs to reduce costs overall, it’s that it needs to reduce SHIPPING costs; etc.) and help you quickly search for the answer choices that deal with that particular gap in logic.

On the downside, if you don’t spend time emphasizing the conclusion, you’re in trouble – burying a conclusion-limiting word or phrase (like “per calorie” or “shipping”) in a long paragraph can be like hiding a needle in a haystack. The Testmaker knows that the untrained are likely to miss these details, and have created trap answers (and just the opportunity to waste time re-reading things that don’t really matter) for those who fall in that group.

5) Scan the Sentence Correction answer choices before you dive into the sentence.
Much like “Answers are Assets” above, a huge help on Sentence Correction problems is to scan the answer choices quickly to see if you can determine where the game is being played (Are they testing pronouns? Verb tenses?). Simply reading a sentence about a strange topic (old excavation sites, a kind of tree that only grows on the leeward slopes of certain mountains…) and looking for anything that strikes you as odd or ungrammatical, that takes time and saps your focus and energy.

However, the GMAT primarily tests a handful of concepts over and over, so if you recognize what is being tested, you can read proactively and look for the words/phrases that directly control that decision you’re being asked to make. Do different answers have different verb tenses? Look for words that signal time (before, since, etc.). Do they involve different pronouns? Read to identify the noun in question and determine which pronoun it needs. You’re not really being tasked with “editing the sentence” as much as your job is to make the proper decision with the choices they’ve already given you. They’ve already narrowed the scope of items you can edit, so identify that scope before you take out the red marking pen across the whole sentence.

6) STOP and avoid rereading.
As the Veritas Prep Reading Comprehension lesson teaches, stop at the end of each paragraph of a reading passage to ask yourself whether you understand Scope, Tone, Organization, and Purpose. The top two time-killers on Reading Comprehension passages/problems are re-reading (you get to the end and realize you don’t really know what you just read) and over-reading (you took several minutes absorbing a lot of details, but now the clock is ticking louder and you haven’t looked at the questions yet).

STOP will help you avoid re-reading (if you weren’t locked in on the first paragraph, you can reread that in 30 seconds and not wait to the end to realize you need to reread the whole thing) and will give you a quick checklist of, “Do I understand just enough to move on?” Details are only important if you’re asked about them, so focus on the major themes (Do you know what the paragraph was about – a quick 5-7 word synopsis is perfect – and why it was written? Good.) and save the details for later.

It may seem ironic that the GMAT is set up to punish hard-workers, but in business, efficiency is everything – the test needs to reward those who work smarter and not just harder, so an effective test day simply cannot be a Labor Day. Use this Labor Day weekend to study effectively so that test day is one on which you prioritize efficiency, not labor.

Would you call yourself a math person? If so, you’ll be glad to know that there are plenty of algebra, geometry, arithmetic, and other types of math problems on the GMAT. Perhaps you like math but need a little review when it comes to the topic of geometry. If so, learn some valuable tips on how to prep for GMAT geometry problems before you get started studying for the exam.

Learn and Practice the Basic Geometry Formulas
Knowing some basic formulas in geometry is an essential step to mastering these questions on the GMAT. One formula you should know is the Pythagorean Theorem, which is a^2 + b^2 = c^2, where c stands for the longest side of a right triangle, while a and b represent the other two sides.

Another formula to remember is the area of a triangle, which is A = 1/2bh, where A is the area, b is the length of the base, and h is the height. The formula for finding the area of a rectangle is l*w = A (length times width equals the area). Once you learn these and other basic geometry formulas for the GMAT, the next step is to put them into practice so you know how to use them when they’re called for on the exam.

Complete Practice Quizzes and QuestionsReviewing problems and their answers and completing GMAT geometry practice questions are two ways to sharpen your skills for this section of the test. This sort of practice also helps you become accustomed to the timing when it comes to GMAT geometry questions. These questions are found within the Quantitative section of the GMAT.

You are given just 75 minutes to finish 37 questions in this section. Of course, not all 37 questions involve geometry – GMAT questions in the Quantitative section also include algebra, arithmetic, and word problems – but working on completing each geometry problem as quickly as possible will help you finish the section within the time limit. In fact, you should work on establishing a rhythm for each section of the GMAT so you don’t have to worry about watching the time.

Use Simple Study Tools to Review Problems
Another way to prepare for GMAT geometry questions is to use study tools such as flashcards to strengthen your skills. Some flashcards are virtual and can be accessed as easily as taking your smartphone out of your pocket. If you prefer traditional paper flashcards, they can also be carried around easily so you can review them during any free moments throughout the day. Not surprisingly, a tremendous amount of review can be accomplished at odd moments during a single day.

In addition, playing geometry games online can help you hone your skills and add some fun to the process at the same time. You could try to beat your previous score on an online geometry game or even compete against others who have played the same game. Challenging another person to a geometry game can sometimes make your performance even better.

Study With a Capable Tutor
Preparing with a tutor can help you to master geometry for GMAT questions. A tutor can offer you encouragement and guide you in your studies. All of our instructors at Veritas Prep have taken the GMAT and earned scores that have put them in the 99th percentile of test-takers. When you study with one of our tutors, you are learning from an experienced instructor as well as someone who has been where you are in the GMAT preparation process.

Our prep courses instruct you on how to approach geometry questions along with every other topic on the GMAT. We know that memorizing facts is not enough: You must apply higher-order thinking to every question, including those that involve geometry. GMAT creators have designed the questions to test some of the skills you will need in the business world.

Taking a practice GMAT gives you an idea of what skills you’ve mastered and which you need to improve. Our staff invites you to take a practice GMAT for free. We’ll give you a score report and a performance analysis so you have a clear picture of what you need to focus on. Then, whether you want help with geometry or another subject on the GMAT, our team of professional instructors is here for you.

One of the interesting things to note about newer GMAC Quant questions is that, while many of these questions test our knowledge of multiples and factors, the phrasing of these questions is often more subtle than earlier versions you might have seen. For example, if I ask you to find the least common multiple of 6 and 9, I’m not being terribly artful about what topic I’m testing you on – the word “multiple” is in the question itself.

But if tell you that I have a certain number of cupcakes and, were I so inclined, I could distribute the same number of cupcakes to each of 6 students with none left over or to each of 9 students with none left over, it’s the same concept, but I’m not telegraphing the subject in the same conspicuous manner as the previous question.

This kind of recognition comes in handy for questions like this one:

All boxes in a certain warehouse were arranged in stacks of 12 boxes each, with no boxes left over. After 60 additional boxes arrived and no boxes were removed, all the boxes in the warehouse were arranged in stacks of 14 boxes each, with no boxes left over. How many boxes were in the warehouse before the 60 additional boxes arrived?

(1) There were fewer than 110 boxes in the warehouse before the 60 additional arrived.
(2) There were fewer than 120 boxes in the warehouse after the 60 additional arrived.

Initially, we have stacks of 12 boxes with no boxes left over, meaning we could have 12 boxes or 24 boxes or 36 boxes, etc. This is when you want to recognize that we’re dealing with a multiple/factor question. That first sentence tells you that the number of boxes is a multiple of 12. After 60 more boxes were added, the boxes were arranged in stacks of 14 with none left over – after this change, the number of boxes is a multiple of 14.

Because 60 is, itself, a multiple of 12, the new number must remain a multiple of 12, as well. [If we called the old number of boxes 12x, the new number would be 12x + 60. We could then factor out a 12 and call this number 12(x + 5.) This number is clearly a multiple of 12.] Therefore the new number, after 60 boxes are added, is a multiple of both 12 and 14. Now we can find the least common multiple of 12 and 14 to ensure that we don’t miss any possibilities.

The prime factorization of 12: 2^2 * 3

The prime factorization of 14: 2 * 7

The least common multiple of 12 and 14: 2^2 * 3 * 7 = 84.

We now know that, after 60 boxes were added, the total number of boxes was a multiple of 84. There could have been 84 boxes or 168 boxes, etc. And before the 60 boxes were added, there could have been 84-60 = 24 boxes or 168-60 = 108 boxes, etc.

A brief summary:

After 60 boxes were added: 84, 168, 252….

Before 60 boxes were added: 24, 108, 192….

That feels like a lot of work to do before even glancing at the statements, but now look at how much easier they are to evaluate!

Statement 1 tells us that there were fewer than 110 boxes before the 60 boxes were added, meaning there could have been 24 boxes to start (and 84 once 60 were added), or there could have been 108 boxes to start (and 168 once 60 were added). Because there are multiple potential solutions here, Statement 1 alone is not sufficient to answer the question.

Statement 2 tells us that there were fewer than 120 boxes after 60 boxes were added. This means there could have been 84 boxes – that’s the only possibility, as the next number, 168, already exceeds 120. So we know for a fact that there are 84 boxes after 60 were added, and 24 boxes before they were added. Statement 2 alone is sufficient, and the answer is B.

Takeaway: questions that look strange or funky are always testing concepts that have been tested in the past – otherwise, the exam wouldn’t be standardized. By making these connections, and recognizing that a verbal clue such as “none left over” really means that we’re talking about multiples and factors, we can recognize even the most abstract patterns on the toughest of GMAT questions.

The Quantitative portion of the GMAT contains questions on a variety of math topics. One of those topics is probability. GMAT questions of this sort ask you to look for the likelihood that something will occur. Probability is not as familiar to many as Algebra, Geometry, and other topics on the test. This is why some test-takers hesitate when they see the word “probability” on a summary of the GMAT. However, this is just another topic that can be mastered with study and practice.

You may already know that there are certain formulas that can help solve GMAT probability questions, but there is more to these problems than teasing out the right answers. Take a look at some advice on how to tackle GMAT probability questions to calm your fears about the test:

Probability Formulas
As you work through GMAT probability practice questions, you will need to know a few formulas. One key formula to remember is that the probability equals the number of desired outcomes divided by the number of possible outcomes. Another formula deals with discrete events and probability – that formula is P(A and B) = P(A)*P(B). Figuring out the probability of an event not occurring is one minus the probability that the event will occur. Putting these formulas into practice is the most effective way to remember them.

Is it Enough to Know the Basic Formulas for Probability?
Some test-takers believe that once you know the formulas related to probability for GMAT questions, then you have the keys to success on this portion of the test. Unfortunately, that is not always the case. The creators of the GMAT are not just looking at your ability to plug numbers into formulas – you must understand what each question is asking and why you arrived at a particular answer. Successful business executives use reason and logic to arrive at the decisions they make. The creators of the GMAT want to see how good you are at using these same tools to solve problems.

The Value of Practice Exams
Taking a practice GMAT can help you determine your skill level when it comes to probability questions and problems on every other section of the test. Also, a practice exam gives you the chance to become accustomed to the amount of time you’ll have to finish the various sections of the test.

At Veritas Prep, we have one free GMAT practice test available to anyone who wants to get an idea of how prepared they are for the test. After you take the practice test, you will receive a score report and thorough performance analysis that lets you know how you fared on each section. Your performance analysis can prove to be one of the most valuable resources you have when starting to prepare for the GMAT. Follow-up practice tests can be just as valuable as the first one you take. These tests reveal your progress on probability problems and other skills on the GMAT. The results can guide you on how to adjust your study schedule to focus more time on the subjects that need it.

Getting the Right Kind of Instruction
When it comes to probability questions, GMAT creators have been known to set subtle traps for test-takers. In some cases, you may happen upon a question with an answer option that jumps out at you as the right choice. This could be a trap.

If you study for the GMAT with Veritas Prep, we can teach you how to spot and avoid those sorts of traps. Our talented instructors have not only taken the GMAT; they have mastered it. Each of our tutors received a score that placed them in the 99th percentile. Consequently, if you study with Veritas Prep, you’ll benefit from the experience and knowledge of tutors who have conquered the GMAT. When it comes to probability questions, GMAT tutors at Veritas Prep have you covered!

In addition to providing you with effective GMAT strategies, tips, and top-quality instruction, we also give you choices regarding the format of your courses. We have prep classes that are given online and in person – learn your lessons where you want, and when you want. You may want to go with our private tutoring option and get a GMAT study plan that is tailored to your needs. Contact Veritas Prep today and dive into your GMAT studies!

In a previous post, we have discussed how to find the total number of factors of a number. What does the total number of factors a number has tell us about that number? One might guess, “Not a lot,” but it actually does tell us quite a bit! If the total number of factors is odd, you know the number must be a perfect square. If the total number of factors is even, you know the number is not a perfect square.

We know that the total number of factors of a number A (prime factorised as X^p * Y^q *…) is given by (p+1)*(q+1)… etc.

So, if we know that a number has, say, 6 total factors, what can we say about the number?

6 = (p+1)*(q+1) = 2*3, so p = 1 and q = 2 or vice versa.

A = X^1 * Y^2 where X and Y are distinct prime numbers.

Today, we will look at a data sufficiency question in which we can use factors to deduce much more information than what we might first guess:

When the digits of a two-digit, positive integer M are reversed, the result is the two-digit, positive integer N. If M > N, what is the value of M?

With this question, we are told that M is a two-digit integer and N is obtained by reversing it. So if M = 21, then N = 12; if M = 83, then N = 38 (keeping in mind that M must be greater than N). In the generic form:

M = 10a + b and N =10b + a (where a and b are single-digit numbers from 1 to 9. Neither can be 0 or greater than 9 since both M and N are two-digit numbers.)

We also know that no matter what M and N are, M > N. Therefore:

10a + b > 10b + a
9a > 9b
a > b

Let’s examine both of our given statements:

Statement 1: The integer (M – N) has 12 unique factors.

First, let’s figure out what M – N is:

M – N = (10a + b) – (10b + a) = 9a – 9b

Say M – N = A. This would mean A = 9(a-b) = 3^2 * (a-b)

The total number of factors of A where A = X^p * Y^q *… can be calculated using the formula (p+1)*(q+1)* …

We know that A has 3^2 as a factor, so X = 3 and p = 2. Therefore, the total number of factors would be (2+1)*(q+1)*… = 3*(q+1)*… = 12, so (q+1)*… must be 4.

Case 1:
This means q may be 3 so that (q+1) is 4. Since a-b must be less than or equal to 9 and must also be the cube of a number, (a-b) must be 8. (Note that a-b cannot be 1 because then the total number of factors of A would only be 3.)

So, a must be 9 and b must be 1 in this case (since a > b). The integers will be 91 and 19, and since M > N, M = 91.

Case 2:
Another possibility is that (a-b) is a product of two prime factors (other than 3), both with the power of 1. In that case, the total number of factors = (2+1)*(1+1)*(1+1) = 12

Note, however, that the two prime factors (other than 3) with the smallest product is 2*5 = 10, but the difference of two single-digit positive integers cannot be 10. This means that only Case 1 can be true, therefore, Statement 1 alone is sufficient. This is certainly not what we expected to find from just the total number of factors!

We get no new information with this statement; (a-b) can be any integer, such as 2 (a = 5, b = 3 or a = 7, b = 5), etc. This statement alone is insufficient, therefore our answer is A.

Don’t take the given data of a GMAT question at face value, especially if you are expecting questions from the 700+ range. Ensure that you have deduced everything that you can from it before coming to a conclusion.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

An individual who is creating a study plan for the GMAT knows that math must be a part of the equation. Though many people love all sorts of math, there are some who become worried about the Quantitative portion of the exam.

If you’re concerned about the math questions on the GMAT, it can be useful to become more familiar with the specific content in this section. Find out about the types of problems in the Quantitative section and consider some GMAT geometry formulas. Also, check out a gathering of tips on how to prep in an effective way:

What is in the Quantitative Section?
Data Sufficiency and Problem-Solving are the two types of questions in the Quantitative section. The Problem-Solving questions are multiple-choice and test your skills in algebra, basic arithmetic, and geometry. The basic arithmetic questions involve decimals, positive and negative integers, fractions, percentages, and averages. The problems you find in this section are on par with the level of material taught in high school math classes. Though many of the questions on the exam involve basic arithmetic, it’s helpful to have a GMAT formula sheet to refer to when preparing for algebra and geometry problems.

GMAT Formulas for the Math Section
Your GMAT math formulas cheat sheet should include the Pythagorean Theorem. This formula helps you to find the measurement of the third side of a right triangle when given the measurements of the other two sides. Another item on your GMAT math cheat sheet should be A = 1/2 bh, which is the formula for finding the area of a triangle. Distance = rate*time is a very helpful formula to know, too. Find the area of a rectangle in fast fashion by using the formula A = lw. The formula A = s2 will help you discover the area of a square.

Moving Beyond Memorization

A GMAT math formulas cheat sheet is an effective study tool, but it’s equally important to know which formula to apply to a problem, so you should spend time practicing problems that employ each of those formulas. This way, on test day, you’ll be familiar with the formulas and feel comfortable using them. The easiest way to do this, of course, is to let us help you.

The expert instructors at Veritas Prep partner with students to help them learn and to practice these formulas for the Quantitative section. We hire tutors who have excellent teaching skills as well as GMAT scores in the 99th percentile. When you study with us, you know you’re learning from the best! Our instructors work through practice math problems with you to ensure that you understand how to solve them in the most efficient way.

Get the Timing Right
Test-takers are given 75 minutes to tackle the 37 questions in the Quantitative section. This sounds like a long time, but if you get hung up on one question for several minutes, you could end up running out of time for this section. In order to avoid this, you should take timed practice tests. Taking timed tests allows you to establish a rhythm for solving problems and answering questions. Once you establish a rhythm, you don’t have to be so concerned about running out of time before you finish all of the problems.

More Tips for Mastering the Quantitative Section
Studying with a GMAT math cheat sheet is one way to prepare for the test. Another way to save test time and make questions more manageable is to eliminate answer options that are clearly wrong – this allows your mind to focus only on the legitimate choices. Estimating the answer to a problem as you read through it is another way to save test time and arrive at answers more quickly.

Our GMAT curriculum teaches you how to approach questions on the separate math topics within the Quantitative section. Our strategies give you the tools you need to problem-solve like a business professional! We are proud to provide both online and in-person courses that prepare you for the GMAT. Veritas Prep instructors offer solid instruction as well as encouragement to individuals with the goal of acing the GMAT and getting into a preferred business school. Let us partner with you on the road to GMAT success! Contact us to talk with one of our course advisers today.

The other day, while working with a tutoring student, I was enumerating the virtues of various test-taking strategies when the student sheepishly interrupted my eloquent paean to picking numbers. She’d read somewhere that these strategies were fine for easy to moderate questions, but that for the toughest questions, you just had to bear down and solve the problem formally. Clearly, she is not a regular reader of our fine blog.

As luck would have it, on her previous practice exam she’d received the following problem, which both illustrates the value of picking numbers and demonstrates why this approach works so well.

A total of 30 percent of the geese included in a certain migration study were male. If some of the geese migrated during the study and 20 percent of the migrating geese were male, what was the ratio of the migration rate for the male geese to the migration rate for the female geese?

[Migration rate for geese of a certain sex = (number of geese of that sex migrating) / (total number of geese of that sex)]

A) 1/4
B) 7/12
C) 2/3
D) 7/8
E) 8/7

This is a perfect opportunity to break out two of my favorite GMAT tools: picking numbers and making charts. So, let’s say there are 100 geese in our population. That means that if 30% are male, we’ll have 30 male geese and 70 females geese, giving us the following chart:

Male

Female

Total

Migrating

Not-Migrating

Total

30

70

100

Now, let’s say 10 geese were migrating. That means that 90 were not migrating. Moreover, if 20 percent of the migrating geese were male, we know that we’ll have 2 migrating males and 8 migrating females, giving us the following:

Male

Female

Total

Migrating

2

8

10

Not-Migrating

Total

30

70

100

(Note that if we wanted to, we could fill out the rest of the chart, but there’s no reason to, especially when we’re trying to save as much time as possible.)

Our migration rate for the male geese is 2/30 or 1/15. Our migration rate for the female geese is 8/70 or 4/35. Ultimately, we want the ratio of the male migration rate (1/15) to the female migration rate (4/35), so we need to simplify (1/15)/(4/35), or (1*35)/(15*4) = 35/60 = 7/12. And we’re done – B is our answer.

My student was skeptical. How did we know that 10 geese were migrating? What if 20 geese were migrating? Or 50? Shouldn’t that change the result? This is the beauty of picking numbers – it doesn’t matter what number we pick (so long as we don’t end up with an illogical scenario in which, say, the number of migrating male geese is greater than the number of total male geese). To see why, watch what happens when we do this algebraically:

Say that we have a total of “t” geese. If 30% are male, we’ll have 0.30t male geese and 0.70t females geese. Now, let’s call the migrating geese “m.” If 20% are male, we’ll have 0.20m migrating males and 0.80m migrating females. Now our chart will look like this:

Male

Female

Total

Migrating

0.20m

0.80m

m

Not-Migrating

Total

0.30t

0.70t

t

The migration rate for the male geese is 0.20m/0.30t or 2m/3t. The migration rate for the female geese is 0.80m/0.70t or 8m/7t. We want the ratio of the male migration rate (2m/3t) to the female migration rate (8m/7t), so we need to simplify (2m/3t)/(8m/7t) = (2m*7t)/(3t * 8m) = 14mt/24mt = 7mt/12mt = 7/12. It’s clear now why the numbers we picked for m and t don’t matter – they cancel out in the end.

Takeaway: We cannot say this enough: the GMAT is not testing your ability to do formal algebra. It’s testing your ability to make good decisions in a stressful environment. So your goal, when preparing for this test, isn’t to become a virtuoso mathematician, even for the toughest questions. It’s to practice the kind of simple creative thinking that will get you to your answer with the smallest investment of your time.

Your first reaction to the title of this post is probably, “I already know my subtraction!” No surprise there. But what is surprising is that our statistics tell us that the following GMAT question – which is nothing extraordinary, but does involve some tricky subtraction – is a 700-level question. That made us decide to write this post. We will discuss this concept along with the question:

The last digit of 12^12 + 13^13 – 14^14 × 15^15 =

(A) 0(B) 1(C) 5(D) 8(E) 9

This is a simple question based on the cyclicity of units digits. There are 3 terms here: 12^12, 13^13 and (14^14)*(15^15). Let’s find the last digit of each of these terms:

12^12
The units digit of 12 is 2.
2 has a cyclicity of 2 – 4 – 8 – 6.
The cycles end at the powers 4, 8, 12 … etc. So, twelve 2’s will end in a units digit of 6.

13^13
The units digit of 13 is 3.
3 has a cyclicity of 3 – 9 – 7 – 1.
A new cycle starts at the powers 1, 5, 9, 13 … etc. So, thirteen 3’s will end in a units digit of 3.

(14^14)*(15^15)
This term is actually the most simple to manage in the case of its units digit – an even number multiplied by a multiple of 5 will end in 0. Also, note that this will be a huge term compared to the other two terms.

This is what our expression looks like when we consider just the units digits of these terms:

(A number ending in 6) + (A number ending in 3) – (A much greater number ending in 0)

Looking at our most basic options, a number ending in 6 added to a number ending in 3 will give us a number ending in 9 (as 3 + 6 = 9). So, the expression now looks like this:

(A number ending in 9) – (A much greater number ending in 0)

It is at this point that many people mess up. They deduce that 9-0 will end in a 9, and hence, the answer will be E. All their effort goes to waste when they do this. Let’s see why:

How do you subtract one number out of another? Take, for example, 10-7 = 3

This can also be written as 7-10 = -3. (Here, you are still subtracting the number with a lower absolute value from the number with a greater absolute value, but giving it a negative sign.)

Let’s try to look at this in tabular form. The number with the greater absolute value goes on the top and the number with the smaller absolute value goes under it. You then subtract and the result gets the sign of the number with the greater absolute value.

So, the number with greater absolute value is always on top. Going back to our original question now, (A number ending in 9) – (A much greater number ending in 0) will look like:

abcd0– pq9
ghjk1

Ignoring the letter variables (these are simply placeholders), note that the greater number ending in 0 will be on the top and the smaller one ending in 9 will be below it. This means the answer will be a negative number ending in a units digit of 1. Therefore, our answer is B.

As we learn more advanced concepts, make sure you are not taking your basic principles for granted!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such asthis blog!

On this Friday, ending the first week of the Rio Olympics, your office has undoubtedly said the name “Simone” exponentially more than ever before. Michael Phelps’ blowout win – his 4th straight – in the 200 IM was incredible, but last night belonged to two Texans named Simone.

Swimmer Simone Manuel and gymnast Simone Biles each won historic gold medals, and if you’re at all inspired to pursue your own “go for the gold” success in business school (maybe Stanford like Manuel, or UCLA like Biles), you can learn a lot from the Olympic experience. Two lessons, in particular, stand out from the performance of Biles and her “Final Five” teammates:

Connect Your Skills
There’s no way to watch Olympic gymnastics and not be overwhelmingly impressed by the skills that each gymnast brings to competition. So at times it’s frustrating and saddening to hear the TV announcers discuss deduction after deduction; shouldn’t everyone at all times just be yelling, “Wow!!!!” at the otherworldly talents of each athlete?

Much like the GMAT, though, Olympic gymnastics is not about the sheer possession of these skills – at that level, everyone has them. It’s more about the ability to execute them and, as becomes evident from the expert commentary of Tim Dagget and Nastia Liukin, to connect them. It’s not the uneven bars handstand or release itself that wins the gold, it’s the ability to connect skill after skill as part of a routine. The line, “She was supposed to connect that skill to another…” is always followed by, “That will be a deduction” – both in Olympic gymnastics and on the GMAT.

How does that affect you?

By test day, you had better have all of the necessary skills to compete on the GMAT Quant Section. Area of a triangle, Pythagorean Theorem, Difference of Squares…if you don’t know these rules, you’re absolutely sunk. But to do really well, you need to quickly connect skill to skill, and connect items in the problems to the skills necessary to work with them. For example:

If a problem includes a term x^4 – 1, you should immediately be thinking, “That connects really well to the Difference of Squares rule: a^2 – b^2 = (a + b)(a – b), and since x^4 is a square [it’s (x^2)^2] and 1 is a square (it’s 1^2), I can write that as (x^2 + 1)(x^2 – 1), and for good measure I could apply Difference of Squares to the (x^2 – 1) term too.” The GMAT won’t ever specifically tell you, “Use the Difference of Squares,” so it’s your job to immediately connect the symptoms of Difference of Squares (an even exponent, a subtraction sign, a square of some kind, even if it’s 1) to the opportunity to use it.

If you see a right triangle, you should recognize that Area and Pythagorean Theorem easily connect. In a^2 + b^2 = c^2, sides a and b are perpendicular and allow you to use them as the base and height in the area formula. And the Pythagorean Theorem includes three squares with the opportunity to create subtraction [you could write it as a^2 = c^2 – b^2, allowing you to say that a^2 = (c + b)(c – b)…], so you could connect yet another skill to it to help solve for variables.

Similarly, if you see a square or rectangle, its diagonal is the hypotenuse of a right triangle, allowing you to use the sides as a and b in the Pythagorean setup, which could also connect to Difference of Squares…etc.

When you initially learned most of these skills in high school (much like when Biles, Aly Raisman, Gabby Douglas, etc. learned handstands and cartwheels in Gymboree), you learned them as individual, isolated skills. “Here’s the formula, and here are 10 questions that test it.” On the GMAT – as in the Olympics – you’re being tested more on your ability to connect them, to see opportunities to use a skill that’s not obvious at first (“Well, I’m not sure what to do but I do have multiple squared terms so let me try to apply Difference of Squares…or maybe I can use a and b in the Area calculation.”), but that helps you build more knowledge of the problem.

So as you study, don’t just learn individual skills. Look for opportunities to connect them, and look for signals that will tell you that a connection is possible. A rectangle problem with a square root of 3 in the answer choices should tell you “the diagonal of this rectangle may very well be connected to a 30-60-90 triangle, since those have the 1, √3, 2 side ratio…” The GMAT is about connections more so than just skills, so study accordingly.

Stick the Landing
If you’re like most in the “every four years I love gymnastics for exactly one week” camp, the single most important thing you look for on any apparatus is, “Did he/she stick the landing?” A hop or a step on the landing is the most noticeable deduction on a gymnastics routine…and the same holds true for the GMAT.

Again, the GMAT is testing you on how well you connect a variety of skills, so naturally there are places for you to finish the problem a step short. A problem that requires you to leverage the Pythagorean Theorem and the Area of a Triangle may ask for the sum of sides A and B, for example, but if you’ve solved for the sides individually first, you might see a particular value (A = 6) on your noteboard and in the answer choices and choose it without double checking that you answered the proper question.

That is a horrible and unnecessary “deduction” on your GMAT score: you did all the work right, all the hard part right (akin to the flip-and-two-twists in the air on your vault or the dazzling array of jumps and handstands on the tiny beam) and then botched the landing.

On problems that include more than one variable, circle the variable that the test is looking for and then make sure that you submit the proper answer for that variable. If a problem asks for a combination of variables (a + b, for example), write that down at the top of your scratchwork and go back to it after you’ve calculated. Take active steps to ensure that you stick the landing, because nothing is worse than doing all the work right and then still getting the problem wrong.

In summary, recognize that there are plenty of similarities between the GMAT and GyMnAsTics [the scoring system is too complex for the layman to worry about, the “Final Five” are more important than you think (hint: the test can’t really use the last five questions of a section for research purposes since so many people are rushing and guessing), etc.]. So take a lesson from Simone Biles and her gold-medal-winning teammates: connect your skills, stick the landing, and you’ll see your score vault to Olympian heights.

We have covered the concepts of direct, inverse and jointvariation in previous posts. Today, we will discuss what we mean by “linearly related”. A linear relation is one which, when plotted on a graph, is a straight line. In linear relationships, any given change in an independent variable will produce a corresponding change in the dependent variable, just like a change in the x-coordinate produces a corresponding change in the y-coordinate on a line.

We know the equation of a line: it is y = mx + c, where m is the slope and c is a constant.

Let’s illustrate this concept with a GMAT question. This question may not seem like a geometry question, but using the concept of linear relations can make it easy to find the answer:

A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?

(A) 20(B) 36(C) 48(D) 60(E) 84

Let’s think of the two scales R and S as x- and y-coordinates. We can get two equations for the line that depicts their relationship:

30 = 6m + c ……. (I)
60 = 24m + c ……(II)

(II) – (I)
30 = 18m
m = 30/18 = 5/3

Plugging m = 5/3 in (I), we get:

30 = 6*(5/3) + c
c = 20

Therefore, the equation is S = (5/3)R + 20. Let’s plug in S = 100 to get the value of R:

100 = (5/3)R + 20
R = 48

48 (answer choice C) is our answer.

Alternatively, we have discussed the concept of slope and how to deal with it without any equations in this post. Think of each corresponding pair of R and S as points lying on a line – (6, 30) and (24, 60) are points on a line, so what will (r, 100) be on the same line?

We see that an increase of 18 in the x-coordinate (from 6 to 24) causes an increase of 30 in the y-coordinate (from 30 to 60).

So, the y-coordinate increases by 30/18 = 5/3 for every 1 point increase in the x-coordinate (this is the concept of slope).

From 60 to 100, the increase in the y-coordinate is 40, so the x-coordinate will also increase from 24 to 24 + 40*(3/5) = 48. Again, C is our answer.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

Students who are taking the GMAT are going to encounter combinatorics problems. If you are a little rusty on your math topics, you may be asking, “What is combinatorics?” Combinatorics has to do with counting and evaluating the possibilities within a scenario that involve various amounts of people or things. Learn more about GMAT combinatorics questions and how to arrive at the right answers to be better prepared for the test.

Permutations
Picture a certain number of people or objects. Permutations are the possible arrangements that those people or objects can be in. One of the things you have to decide when looking at combinatorics problems is whether order is an important factor. If order is important in a problem, then the answer has to do with permutations. If order is not important in a problem, then the answer deals with combinations.

For example, say you line up five postcards from different cities on a tabletop. You may wonder how many different orders you can put these postcards in. Another way to say that would be, “How many different permutations can I make with these five postcards?” To figure out this problem, you would need the help of an equation: 5! = (5) (4) (3) (2) (1) = 120. The exclamation point in the formula is a symbol that means “factorial.”

Combinations
When working on combinatorics questions that deal with combinations, the order/arrangement of items is not important. For example, say that you have eight books and you want to know how many ways you can group three of those books on a library shelf. You could plug numbers into the three places in this formula to figure out the answer: (8) (7) (6) = 336 ways. This is the slot method of solving a combination problem.

Combinations With a Large Amount of Numbers
You will quickly find yourself needing combinatorics help if you try to count up a lot of numbers in one combination problem on the GMAT. Furthermore, you’ll use a lot of valuable test time with this counting method. Knowing the formula for combinations can help you to find the solution to a problem in a much shorter amount of time. The formula is nCr = n!/r!(n-r)! Here, n is the total number of options, r is the number of options chosen, and ! is the symbol for factorial.

Preparing for Applied Combinatorics Questions on the GMAT
One of the most effective ways of preparing for applied combinatorics questions is to take practice tests and review the various steps of problems. You want to get into the habit of approaching a problem by asking yourself whether order is a factor in a problem. This will help you determine whether a problem deals with permutations or combinations. Then, you can start to attack a problem from the right angle.

In addition, it’s important to time yourself when taking a practice Quantitative test. Though there are not many of these problems on the test, you have to get into the habit of spending only a certain amount of minutes on each problem so you don’t run out of test time before finishing.

We have a program of studyat Veritas Prep that prepares you for questions on combinatorics as well as all of the other problems in the Quantitative section. We instruct you on how to approach test questions instead of just coaching you on how to memorize facts. Pair up with one of our skilled instructors at Veritas Prep and you will be studying with someone who scored in the 99th percentile on the GMAT. We believe that in order to perform at your best on the GMAT, you have to learn from a first-rate instructor! Our instructors can work through a combinatorics tutorial with you to determine what your strengths and weaknesses are in this branch of math. Then, we give you strategies that help you to improve.

For your convenience, we offer both in-person and online GMAT prep courses. We recognize that professionals in the business world have busy schedules, so we provide several study options to fit your life. When it comes to the topic of combinatorics, GMAT tips, instruction, and encouragement, we are your test prep experts. Contact us today and let us know how we can help you achieve your top GMAT score!

If you’ve been paying attention to the exciting world of GMAT prep, you know thatGMAC released two new practice tests fairly recently. I’d mentioned in a previous post that I was going to write about any conspicuous trends I noted, and one unmistakable pattern I’ve seen with my students is that probability questions seem to be cropping up with greater and greater frequency.

While these questions don’t seem fundamentally different from what we’ve seen in the past, there does seem to be a greater emphasis on probability questions for which a strong command of combinations and permutations will prove indispensable.

First, recall that the probability of x is the number ways x can occur/number of total possible outcomes (or p(x) = # desired/ # total). Another way to think about this equation is to see it as a ratio of two combinations or permutations. The number of ways x can occur is one combination (or permutation), and the total number of possible outcomes is another.

Keeping this in mind, let’s tackle this new official prompt:

From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?

(A) 3/7(B) 5/12(C) 27/70(D) 2/7(E) 9/35

Typically, I’ll start by calculating the total number of possible outcomes, as this calculation tends to be the more straightforward one. We’ve got 8 volunteers, and we want to know the number of total ways we can select 4 people from these 8 volunteers. Note, also, that the order does not matter – group of Tiffany, Mike, Louis, and Amy is the same as a group of Louis, Amy, Mike, and Tiffany. We’re not assigning titles or putting anyone in seats, so this is a combination.

If we use our combination formula N!/[(K!*(N-K)!] then N, our total pool of candidates, is 8, and K, the number we’re selecting, is 4. We get 8!/(4!*4!), which comes out to 70. At this point, we know that the denominator must be a factor of 70, so anything that doesn’t meet this criterion is out. In this case, this only allows us to eliminate B.

Now we want our desired outcomes, in which Andrew is selected and Karen is not. Imagine that you’re responsible for assembling this group of four from a total pool of eight people. You plan on putting your group of four in a conference room. Your supervisor tells you that Andrew must be in and Karen must not be, so you take Andrew and put him in the conference room. Now you’ve got three more spots to fill and seven people remaining. But remember that Karen cannot be part of this group. That means you only have 6 people to choose from to fill those other 3 spots in the room.

Put another way, think of the combination as the number of choices you have. Andrew and Karen are not choices – you’ve been ordered to include one of them and not the other. Of the 4 spots in the conference room, you only get to choose 3. And you’re only selecting from the other 6 people for those spots. Now N = 6 and K = 3. Plugging these into our trusty combination formula, we get 6!/(3!*3!), which comes out to 20.

Summarizing, we know that there are 20 ways to create our desired group of 4, and 70 total ways to select 4 people from a pool of 8, giving us a probability of 20/70, or 2/7, so the correct answer is D.

Takeaway: Probability questions can be viewed as ratios of combinations or permutations, so when you brush up on combinatorics, you’re also bolstering your probability fundamentals. Anytime you’re stuck on a complex probability question, break your calculation down into its component parts – find the total number of possible outcomes first, then find the total number of desired outcomes. Like virtually every hard question on the GMAT, probability questions are never as hard as they first seem.

Most people feel that the topic of number properties is hard or at least a little tricky. The reason is that no matter how much effort you put into it, you will still come across new concepts every time you sit with some 700+ level problems of this topic. There will be some concepts you don’t know and will need to “figure out” during the actual test. I came across one such question the other day. It brought forth a concept I hadn’t thought about before so I decided to share it today:

Say you have N consecutive integers (starting from any integer). What can you say about their sum? What can you say about their product?

Say N = 3
The numbers are 5, 6, 7 (any three consecutive numbers)
Their sum is 5 + 6 + 7 = 18
Their product is 5*6*7 = 210
Note that both the sum and the product are divisible by 3 (i.e. N).

Say N = 4
The numbers are 3, 4, 5, 6 (any five consecutive numbers)
Their sum is 3 + 4 + 5 + 6 = 18
Their product is 3*4*5*6 = 360
Now note that the sum is not divisible by 4, but the product is divisible by 4.

If N is odd then the sum of N consecutive integers is divisible by N, but this is not so if N is even.
Why is this so? Let’s try to generalize – if we have N consecutive numbers, they will be written in the form:

1 and 3 add up to give 4 but we still have a 2 extra. So the sum of four consecutive integers will not be a multiple of 4.

Let’s now consider the product of N consecutive integers.

In any N consecutive integers, there will be a multiple of N. Hence, the product will always be a multiple of N.

Now take a quick look at the GMAT question that brought this concept into focus:

Which of the following must be true?1) The sum of N consecutive integers is always divisible by N.2) If N is even then the sum of N consecutive integers is divisible by N.3) If N is odd then the sum of N consecutive integers is divisible by N.4) The Product of K consecutive integers is divisible by K.5) The product of K consecutive integers is divisible by K!

(A) 1, 4, 5(B) 3, 4, 5(C) 4 and 5(D) 1, 2, 3, 4(E) only 4

Let’s start with the first three statements this question gives us. We can see that out of Statements 1, 2 and 3, only Statement 3 will be true for all acceptable values of N. Therefore, all the answer choices that include Statements 1 and 2 are out, i.e. options A and D are out. The answer choices that don’t have Statement 3 are also out, i.e. options C and E are out. This leaves us with only answer choice B, and therefore, B is our answer.

This question is a direct application of what we learned above so it doesn’t add much value to our learning as such, but it does have an interesting point. By establishing that B is the answer, we are saying that Statement 5 must be true.

5) The product of K consecutive integers is divisible by K!

We will leave it to you to try to prove this!

(For more advanced number properties on the GMAT, check out Parts I, II, III, IV and V of this series.)

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

48 is completely divisible by 8, so we conclude that 65748048 is also divisible by 8.

A valid question here is, “What about the remaining five digits? Why do we ignore them?”

Breaking down M, we can see that 65748048 = 65748000 + 048 (we’ve separated the last three digits).

Now note that 65748000 = 65748 * 1000. Since 1000 has three 0s, it is made up of three 2s and three 5s. Because 1000 it has three 2s as factor, it also has 8 as a factor. This means 65748000 has 8 as a factor by virtue of its three 0s.

All we need to worry about now is the last three digits, 048. If this is divisible by 8, 65748048 will also be divisible by 8. If it is not, 65748048 will not be divisible by 8.

In case the last three digits are not divisible by 8, you can still find the remainder of the number. Whatever remainder you get after dividing the last three digits by 8 will be the remainder when you divide the entire number by 8. This should not be a surprise to you now – 65748000 won’t have a remainder when divided by 8 since it is divisible by 8, so whatever the remainder is when the last 3 digits are divided by 8 will be the remainder when the entire number is divided by 8.

In the generic case, the number M will be split into a number with n zeroes and another number with n digits. The number with n zeroes will be divisible by 2^n because it has n 2s as factors. We just need to see the divisibility of the number with n digits.

We hope you have understood this concept. Let’s take look at a quick GMAT question to see this in action:

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches theGMAT for Veritas Prep and regularly participates in content development projects such as this blog!

During the first session of each new class I teach, we do a quick primer on the utility of units digits. Imagine I want to solve 130,467 * 367,569. Without a calculator, we are surely entering a world of hurt. But we can see almost instantaneously what the units digit of this product would be.

The units digit of 130,467 * 367,569 would be the same as the units digit of 7*9, as only the units digits of the larger numbers are relevant in such a calculation. 7*9 = 63, so the units digit of 130,467 * 367,569 is 3. This is one of those concepts that is so simple and elegant that it seems too good to be true.

And yet, this simple, elegant rule comes into play on the GMAT with surprising frequency.

Take this question for example:

If n is a positive integer, how many of the ten digits from 0 through 9 could be the units digit of n^3?

A) threeB) fourC) sixD) nineE) ten

Surely, you think, the solution to this question can’t be as simple as cubing the easiest possible numbers to see how many different units digits result. And yet that’s exactly what we’d do here.

1^3 = 1

2^3 = 8

3^3 = 27 à units 7

4^3 = 64 à units 4

5^3 = ends in 5 (Fun fact: 5 raised to any positive integer will end in 5.)

6^3 = ends in 6 (Fun fact: 6 raised to any positive integer will end in 6.)

7^3 = ends in 3 (Well 7*7 = 49. 49*7 isn’t that hard to calculate, but only the units digit matters, and 9*7 is 63, so 7^3 will end in 3.)

8^3 = ends in 2 (Well, 8*8 = 64, and 4*8 = 32, so 8^3 will end in 2.)

9^3 = ends in 9 (9*9 = 81 and 1 * 9 = 9, so 9^3 will end in 9.)

10^3 = ends in 0

Amazingly, when I cube all the integers from 1 to 10 inclusive, I get 10 different units digits. Pretty neat. The answer is E.

Of course, this question specifically invoked the term “units digit.” What are the odds of that happening? Maybe not terribly high, but any time there’s a painful calculation, you’d want to consider thinking about the units digits.

Take this question, for example:

A certain stock exchange designates each stock with a one, two or three letter code, where each letter is selected from the 26 letters of the alphabet. If the letters may be replaced and if the same letters used in a different order constitute a different code, how many different stocks is it possible to uniquely designate with these codes?

A) 2,951
B) 8,125
C) 15,600
D) 16,302
E) 18,278

Conceptually, this one doesn’t seem that bad.

If I wanted to make a one-letter code, there’d be 26 ways I could do so.

If I wanted to make a two-letter code, there’d be 26*26 or 26^2 ways I could do so.

If I wanted to make a three-letter code, there’d be 26*26*26, or 26^3 ways I could so.

So the total number of codes I could make, given the conditions of the problem, would be 26 + 26^2 + 26^3. Hopefully, at this point, you notice two things. First, this arithmetic will be deeply unpleasant to do. Second, all of the answer choices have different units digits!

Now remember that 6 raised to any positive integer will always end in 6. So the units digit of 26 is 6, and the units digit of 26^2 is 6 and the units digit of 26^3 is also 6. Therefore, the units digit of 26 + 26^2 + 26^3 will be the same as the units digit of 6 + 6 + 6. Because 6 + 6 + 6 = 18, our answer will end in an 8. The only possibility here is E. Pretty nifty.

Takeaway: Painful arithmetic can always be avoided on the GMAT. When calculating large numbers, note that we can quickly find the units digit with minimal effort. If all the answer choices have different units digits, the question writer is blatantly telegraphing how to approach this problem.

Fans of The Big Bang Theory will remember Sheldon Cooper’s quote from an old episode on his favorite number:

“The best number is 73. Why? 73 is the 21st prime number. Its mirror, 37, is the 12th and its mirror, 21, is the product of multiplying 7 and 3… and in binary 73 is a palindrome, 1001001, which backwards is 1001001.”

Though Sheldon’s logic is infallible, my favorite number is 1001 because it has a special role in standardized tests.

1001 is 1 more than 1000 and hence, is sometimes split as (1000 + 1). It sometimes appears in the a^2 – b^2 format such as 1001^2 – 1, and its factors are 7, 11 and 13 (not the factors we usually work with).

Due to its unusual factors and its convenient location (right next to 1000), it could be a part of some tough-looking GMAT questions and should be remembered as a “special” number. Let’s look at a question to understand how to work with this number.

Which of the following is a factor of 1001^(32) – 1 ?

(A) 768(B) 819(C) 826(D) 858(E) 924

Note that 1001 is raised to the power 32. This is not an exponent we can easily handle. If we try to use a binomial here and split 1001 into (1000 + 1), all we will achieve is that upon expanding the given expression, 1 will be cancelled out by -1 and all other terms will have 1000 in common. None of the answer choices are factors of 1000, however, so we must look for some other factor of 1001^(32) – 1.

Without a calculator, it is not possible for us to find the factors of 1001^(32) – 1, but we do know the prime factors of 1001 and hence, the prime factors of 1001^32. We may not be able to say which numbers are factors of 1001^(32) – 1, but we will be able to say which numbers are certainly not factors of this!

Now, what can we say about the prime factors of 1001^(32) – 1? Whatever they are, they are certainly not 7, 11 or 13 – two consecutive integers cannot have any common prime factor (discussed here and continued here).

Now look at the answer choices and try dividing each by 7:

(A) 768 – Not divisible by 7

(B) 819 – Divisible by 7

(C) 826 – Divisible by 7

(D) 858 – Not divisible by 7

(E) 924 – Divisible by 7

Options B, C and E are eliminated. They certainly cannot be factors of 1001^(32) – 1 since they have 7 as a prime factor, and we know 1001^(32) – 1 cannot have 7 as a prime factor.

Now try dividing the remaining options by 11:

(A) 768 – Not divisible by 11

(D) 858 – Divisible by 11

D can also be eliminated now because it has 11 as a factor. By process of elimination, the answer is A; it must be a factor of 1001^(32) – 1.

I hope you see how easily we used the factors of 1001 to help us solve this difficult-looking question. And yes, another attractive feature of 1001 – it is a palindrome in the decimal representation itself!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

In seemingly the most important development in world history since humans learned to create fire, Pokemon Go has arrived and is taking the world by storm. Rivaling Twitter and Facebook for mobile phone attention and battling the omnipresent selfie as a means of death-by-mobile-phone, Pokemon Go is everywhere you want to be…and often in places you don’t.

And that is why Pokemon Go is responsible for an ever-important GMAT lesson.

Perhaps most newsworthy about Pokemon Go these days is the dangerous and improper places that it has led its avid users. On the improper side, such solemn and dignified places as the national Holocaust Museum and Arlington National Cemetery have had to actively prohibit gamersfrom descending upon mourners/commemorators while playing the game. And as for danger, there have been several instances of thieves luring gamers into traps and therefore robbing them of valuable (if you’re playing the game, you definitely have a smartphone) items.

And the GMAT can and will do the same thing.

How?

If you’re reading this on our GMAT blog, you’ve undoubtedly already learned that, on Data Sufficiency problems, you cannot assume that a variable is positive, or that it is an integer. But think about what makes Pokemon Go users so vulnerable to being lured into a robbery or to losing track of basic human decency. They’re so invested in the game that they lose track of the situations they’re being lured into.

Similarly, the most dangerous GMAT traps are those for which you should absolutely know better, but the testmaker has gotten your mind so invested in another “game” that you lose track of something basic. Consider the example:

If y is an odd integer and the product of x and y equals 222, what is the value of x?

(1) x is a prime number (2) y is a 3 digit number

Statement 1 is clearly sufficient. Since y is odd, and an integer, and the product of integers x and y is an even integer, that means that x must be even. And since x also has to be prime (which is how you know it’s an integer, too), the only even prime is 2, making x = 2.

From there your mind is fixated on the game. You can quickly see that in that case y = 111 and x = 2. Which you then have to forget about as you attack Statement 2. But here’s the reason that less than 25% of users in the Veritas Prep Question Bank get this right, while nearly half incorrectly choose D. Statement 1 has gotten your mind fixated on the even/odd/prime game, meaning that you may only be thinking about integers (and positive integers at that) at this point.

That y is a 3-digit number DOES NOT mean that it has to be 111. It could be -111 (making x = -2) or 333 (making x = 2/3). So only Statement 1 alone is sufficient, but the larger lesson is more important. Just like Pokemon Go has the potential to pollute your mind and have you see the real world through its “enhanced reality” lens, so does a statement that satisfies your intellect (“Ah, 2 is the only even prime number!”) give you just enough tunnel vision that you make poor decisions and fall for traps.

The secret here is that almost no one scoring above a 500 carries over all of Statement 1 (“Oh, well I already know that x = 2!”) – a total rookie mistake. It’s that Statement 1 got you fixated on definitions of types of integers (prime, even, odd) and therefore got your mind looking through the “enhanced reality” of integers-only.

The lesson? Much like Pokemon Go, the GMAT has tools to get you so invested in a particular facet of a game that you lose your universal awareness of your surroundings. Know that going in – that you have to consciously step back from that enhanced reality you’ve gained after Statement 1 and look at the whole picture. So take a lesson from Pokemon Go and know when to stop and step back.

My students have a hard time understanding what makes a difficult GMAT question difficult. They assume that the tougher questions are either testing something they don’t know, or that these problems involve a dizzying level of complexity that requires an algebraic proficiency that’s simply beyond them.

One of my main goals in teaching a class is to persuade everyone that this is not, in fact, how hard questions work on this test. Hard questions don’t ask you do to something you don’t know how to do. Rather, they’re cleverly designed to provoke an anxiety response that makes it difficult to realize that you do know exactly how to solve the problem.

Take this official question, for example:

Let a, b, c and d be nonzero real numbers. If the quadratic equation ax(cx + d) = -b(cx +d) is solved for x, which of the following is a possible ratio of the 2 solutions?

A) –ab/cdB) –ac/bdC) –ad/bcD) ab/cdE) ad/bc

Most students see this and panic. Often, they’ll start by multiplying out the left side of the equation, see that the expression is horrible (acx^2 + adx), and take this as evidence that this question is beyond their skill level. And, of course, the question was designed to elicit precisely this response. So when I do this problem in class, I always start by telling my students, much to their surprise, that every one of them already knows how to do this. They’ve just succumbed to the question writer’s attempt to convince them otherwise.

So let’s start simple. I’ll write the following on the board: xy = 0. Then I’ll ask what we know about x or y. And my students shrug and say x or y (or both) is equal to 0. They’ll also wonder what on earth such a simple identity has to do with the algebraic mess of the question they’d been struggling with.

I’ll then write this: zx + zy = 0. Again, I’ll ask what we know about the variables. Most will quickly see that we can factor out a “z” and get z(x+y) = 0. And again, applying the same logic, we see that one of the two components of the product must equal zero – either z = 0 or x + y = 0.

Next, I’ll ask if they would approach the problem any differently if I’d given them zx = -zy – they wouldn’t.

Now it clicks. We can take our initial equation in the aforementioned problem: ax(cx +d) = -b(cx+d), and see that we have a ‘cx + d’ on both sides of the equation, just as we’d had a “z” on both sides of the previous example. If I’m able to get everything on one side of the equation, I can factor out the common term.

Now ax(cx +d) = -b(cx+d) becomes ax(cx +d) + b(cx+d) = 0.

Just as we factored out a “z” in the previous example, we can factor out “cx + d” in this one.

Now we have (cx + d)(ax + b) = 0.

Again, if we multiply two expressions to get a product of zero, we know that at least one of those expressions must equal 0. Either cx + d = 0 or ax + b = 0.

If cx + d = 0, then x = -d/c.

If ax + b = 0, then x = -b/a.

Therefore, our two possible solutions for x are –d/c and –b/a. So, the ratio of the two would simply be (-d/c)/(-b/a). Recall that dividing by a fraction is the equivalent of multiplying by the reciprocal, so we’re ultimately solving for (-d/c)(-a/b). Multiplying two negatives gives us a positive, and we end up with da/cb, which is equivalent to answer choice E.

Takeaway: Anytime you see something on the GMAT that you think you don’t know how to do, remind yourself that the question was designed to create this false impression. You know how to do it – don’t hesitate to dive in and search for how to apply this knowledge.

Both a test-taker at the 48 level and one at the 51 level in the GMAT Quant section, are conceptually strong – given an unlimited time frame, both will be able to solve most GMAT questions correctly. The difference lies in the two things a test-taker at the 51 level does skillfully:

Uses holistic, big-picture methods to solve Quant questions.

Handles questions he or she finds difficult in a timely manner.

We have been discussing holistic methods on this blog for a long time now and will continue discussing them. (Before you continue reading, be sure to check out parts I, II, III, IV, V and VI of this series.)

Today we will focus on “handling the hard questions in a timely manner.” Note that we do not say “solving the hard questions in a timely manner.” Occasionally, one might be required to make a quick call and choose to guess and move on – but again, that is not the focus of this post. We are actually going to talk about the “lightbulb” moment that helps us save on time. There are many such moments for the 51 level test-taker – in fact, the 51 scorers often have time left over after attempting all these questions.

Test takers at the 48 level will also eventually reach the same conclusions but might need much more time. That will put pressure on them the next time they look at the ticking clock, and once their cool is lost, “silly errors” will start creeping in. So it isn’t about just that one question – one can end up botching many other questions too.

There are many steps that can be easily avoided by a lightbulb moment early on. This is especially true for Data Sufficiency questions.

Let’s take an official example:

Pam owns an inventory of unopened packages of corn and rice, which she has purchased for $17 and $13 per package, respectively. How many packages of corn does she have ?

Statement 1: She has $282 worth of packages.

Statement 2: She has twice as many packages of corn as of rice.

A high scorer will easily recognize that this question is based on the concept of “integral solutions to an equation in two variables.” Since, in such real world examples, x and y cannot be negative or fractional, these equations usually have a finite number of solutions.

After we find one solution, we will quickly know how many solutions the equation has, but getting the first set of values that satisfy the equation requires a little bit of brute force.

The good thing here is that this is a Data Sufficiency question – you don’t need to find the actual solution. The only thing we need is to establish that there is a single solution only. (Obviously, there has to be a solution since Pam does own $282 worth of packages.)

This is where the 51 level scorer stops because they never lose sight of the big picture. The “lightbulb” switches on, and now he or she knows that there will be only one set of values that can satisfy this equation. Why? Because y will be less than 17 in the first set of values that satisfies this equation. So if we want to get the next set that satisfies, we will need to subtract y by 17 (and add 13 to x), which will make y negative.

So in any case, there will be a unique solution to this equation. We don’t actually need to find the solution and hence, nothing will be gained by continuing these calculations. Statement 1 is sufficient.

Statement 2: She has twice as many packages of corn as of rice.

Statement 2 gives us no information on the total number of packages or the total amount spent. Hence, we cannot find the total number of packages of corn using this information alone. Therefore, our answer is A.

I hope you see how you can be alert to what you want to handle these Quant questions in a timely manner.

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMATfor Veritas Prep and regularly participates in content development projects such as this blog!

For most of our lives, we ask and answer relatively generic questions: “How’s it going?” “What are you up to this weekend?” “What time do the Cubs play tonight?”

And think about it, what if those questions were more specific: “Are you in a melancholy mood today?” “Are you and Josh going to dinner at Don Antonio’s tonight and ordering table-side guacamole?” “Do the Cubs play at 7:05 tonight on WGN?” If someone is asking those questions instead, you’re probably a bit suspicious. Why so specific? What’s your angle?

The same is true on the GMAT. Most of the question stems you see are relatively generic: “What is the value of x?” “Which of the following would most weaken the author’s argument?” So when the question stem get a little too specific, you should become a bit suspicious. What’s the test going for there? Why so specific?

The overly-specific Critical Reasoning question stem is a great example. Consider the problem:

Raisins are made by drying grapes in the sun. Although some of the sugar in the grapes is caramelized in the process, nothing is added. Moreover, the only thing removed from the grapes is the water that evaporates during the drying, and water contains no calories or nutrients. The fact that raisins contain more iron per food calorie than grapes do is thus puzzling.

Which one of the following, if true, most helps to explain why raisins contain more iron per calorie than do grapes?

(A) Since grapes are bigger than raisins, it takes several bunches of grapes to provide the same amount of iron as a handful of raisins does. (B) Caramelized sugar cannot be digested, so its calories do not count toward the food calorie content of raisins. (C) The body can absorb iron and other nutrients more quickly from grapes than from raisins because of the relatively high water content of grapes. (D) Raisins, but not grapes, are available year-round, so many people get a greater share of their yearly iron intake from raisins than from grapes. (E) Raisins are often eaten in combination with other iron-containing foods, while grapes are usually eaten by themselves.

Look at that question stem: a quick scan naturally shows you that you need to explain/resolve a paradox, but the question goes into even more detail for you. It reaffirms the exact nature of the paradox – it’s not about “iron,” but instead that that raisins contain more iron per calorie than grapes do. By adding that extra description into the question stem, the testmaker is practically yelling at you, “Make sure you consider calories…don’t just focus on iron!” And therefore, you should be prepared for the correct answer B, the only one that addresses calories, and deftly avoid answers A, C, D, and E, which all focus only on iron (and do so tangentially to the paradox).

Strategically speaking, if a Critical Reasoning question stem gets overly specific, you should pay particular attention to the specificity there…it’s most likely directing you to the operative portion of the argument.

Overly specific questions are most helpful in Data Sufficiency questions (and that same logic will help on Problem Solving too, as you’ll see). The testmaker knows that you’ve trained your entire algebraic life to solve for individual variables. So how can a question author use that lifetime of repetition against you? By asking you to solve for a specific combination that doesn’t require you to find the individual values. Consider this example, which appears courtesy the Official Guide for GMAT Quantitative Review:

If x^2 + y^2 = 29, what is the value of (x – y)^2?

(1) xy = 10 (2) x = 5

Two major clues should stand out to you that you need to Leverage Assets on this problem. For one, using both statements together (answer choice C) is dead easy. If xy = 10 and x = 5 then y = 2 and you can solve for any combination of x and y that anyone could ever ask for. But secondly and more subtly, the question stem should jump out as a classic way-too-specific, Leverage Assets question stem. They asked for a really, really specific value: (x – y)^2.

Now, immediately upon seeing that specificity you should be thinking, “That’s too specific…there’s probably a way to solve for that exact value without getting x and y individually.” That thought process alone tells you where to spend your time – you want to really leverage Statement 1 to try to make it work alone.

And if you’re still unconvinced, consider what the specificity does: the “squared” portion removes the question of negative vs. positive from the debate, removing one of the most common reasons that a seemingly-sufficient statement just won’t work. And, furthermore, the common quadratic (x – y)^2 shares an awful lot in common with the x^2 and y^2 elsewhere in the question stem. If you expand the parentheses, you have “What is x^2 – 2xy + y^2?” meaning that you’re already 2/3 of the way there (so to speak), since they’ve spotted you the sum x^2 + y^2.

The important strategy here is that the overly-specific question stem should scream “LEVERAGE ASSETS” and “You don’t need to solve for x and y…there’s probably a way to solve directly for that exact combination.” Since you know that you’re solving for the expanded x^2 – 2xy + y^2, and you already know that x^2 + y^2 = 29, you’re really solving for 29 – 2xy. Since you know from Statement 1 that xy = 20, then 29 – 2xy will be 29 – 2(10), which is 9.

Statement 1 alone is sufficient, even though you don’t know what x and y are individually. And one of the major signals that you should recognize to help you get there is the presence of an overly specific question stem.

So remember, in a world of generic questions, the oddly specific question should arouse a bit of suspicion: the interrogator is up to something! On the GMAT, you can use that to your advantage – an overly specific Critical Reasoning question usually tells you exactly which keywords are the most important, and an overly specific Data Sufficiency question stem begs for you to leverage assets and find a way to get the most out of each statement.

When I was a child, I was terrified of riptides. Partially, this was a function of having been raised by unusually neurotic parents who painstakingly instilled this fear in me, and partially this was a function of having inherited a set of genes that seems to have predisposed me towards neuroticism. (The point, of course, is that my parents are to blame for everything. Perhaps there is a better venue for discussing these issues.)

If there’s a benefit to fears, it’s that they serve as potent motivators to find solutions to the troubling predicaments that prompt them. The solution to dealing with riptides is to avoid struggling against the current. The water is more powerful than you are, so a fight is a losing proposition – rather, you want to wait for an opportunity to swim with the current and allow the surf to bring you back to shore. There’s a profound wisdom here that translates to many domains, including the GMAT.

In class, whenever we review a strategy, my students are usually comfortable applying it almost immediately. Their deeper concern is about when to apply the strategy, as they’ll invariably find that different approaches work with different levels of efficacy on different problems. Moreover, even if one has a good strategy in mind, the way the strategy is best applied is often context-dependent. When we’re picking numbers, we can say that x = 2 or x = 100 or x = 10,000; the key is not to go in with a single approach in mind. Put another way, don’t swim against the arithmetic currents.

Let’s look at some questions to see this approach in action:

At a picnic there were 3 times as many adults as children and twice as many women as men. If there was a total of x men, women, and children at the picnic, how many men were there, in terms of x?

A) x/2B) x/3C) x/4D) x/5E) x/6

The moment we see “x,” we can consider picking numbers. The key here is contemplating how complicated the number should be. Swim with the current – let the question tell you. A quick look at the answer choices reveals that x could be something simple. Ultimately, we’re just dividing this value by 2, 3, 4, 5, or 6.

Keeping this in mind, let’s think about the first line of the question. If there are 3 times as many adults as children, and we’re keeping things simple, we can say that there are 3 adults and 1 child, for a total of 4 people. So, x = 4.

Now, we know that among our 3 adults, there are twice as many women as men. So let’s say there are 2 women and 1 man. Easy enough. In sum, we have 2 women, 1 man, and 1 child at this picnic, and a total of 4 people. The question is how many men are there? There’s just 1! So now we plug x = 4 into the answers and keep going until we find x = 1. Clearly x/4 will work, so C is our answer. The key was to let the question dictate our approach rather than trying to impose an approach on the question.

Let’s try another one:

Last year, sales at Company X were 10% greater in February than in January, 15% less in March than in February, 20% greater in April than in March, 10% less in May than in April, and 5% greater in June than in May. On which month were sales closes to the sales in January?

A) FebruaryB) MarchC) AprilD) MayE) June

Great, you say. It’s a percent question. So you know that picking 100 is often a good idea. So, let’s say sales in January were 100. If we want the month when sales were closest to January’s level, we want the month when sales were closest to 100, Sales in February were 10% greater, so February sales were 110. (Remember that if sales increase by 10%, we can multiply the original number by 1.1. If they decrease by 10% we could multiply by 0.9, and so forth.)

So far so good. Sales in March were 15% less than in February. Well, if sales in Feb were 110, then the sales in March must be 110*(0.85). Hmm… A little tougher, but not insurmountable. Now, sales in April were 20% greater than they were in March, meaning that April sales would be 110*(0.85)*1.2. Uh oh. Once you see that sales are 10% less in May than they were in April, we know that sales will be 110*(0.85)*1.2*0.9.

Now you need to stop. Don’t swim against the current. The arithmetic is getting hard and is going to become time-consuming. The question asks which month is closest to 100, so we don’t have to calculate precise values. We can estimate a bit. Let’s double back and try to simplify month by month, keeping things as simple as possible.

Our February sales were simple: 110. March sales were 110*0.85 – an unpleasant number. So, let’s try thinking about this a little differently. 100*0.85 = 85. 10*0.85 = 8.5. Add them together and we get 85 + 8.5 = 93.5. Let’s make life easier on ourselves – we’ll round up, and call this number 94.

April sales are 20% more than March sales. Well, 20% of 100 is clearly 20, so 20% of 94 will be a little less than that. Say it’s 18. Now sales are up to 94 + 18 = 112. Still not close to 100, so we’ll keep going.

May sales are 10% less than April sales. 10% of 112 is about 11. Subtract 11 from 112, and you get 101. We’re looking for the number closest to 100, so we’ve got our answer – it’s D, May.

Takeaway: Don’t try to impose your will on GMAT questions. Use the structural clues of the problems to dictate how you implement your strategy, and be prepared to adjust midstream. The goal is never to conquer the ocean, but rather, to ride the waves to calmer waters.

One of the most confounding aspects of the GMAT is its tendency to make simple concepts seem far more complex than they are in reality. Percent questions are an excellent example of this.

When I introduce this topic, I’ll typically start by asking my class the following question: If you’ve completed 10% of a project how much is left to do? I have never, in all my years of teaching, had a class that was unable to tell me that 90% of the project remains. It’s more likely that they’ll react as though I’m insulting their collective intelligence. And yet, when test-takers see this concept under pressure, they’ll often fail to recognize it.

Take the following question, for example:

Dara ran on a treadmill that had a readout indicating the time remaining in her exercise session. When the readout indicated 24 min 18 sec, she had completed 10% of her exercise session. The readout indicated which of the following when she had completed 40% of her exercise session.

Hopefully, you’ve noticed that this question is testing the same simple concept that I use when introducing percent problems to my class. And yet, in my experience, a solid majority of students are stumped by this problem. The reason, I suspect, is twofold. First, that figure – 24 min. 18 sec. – is decidedly unfriendly. Painful math often lends itself to careless mistakes and can easily trigger a panic response. Second, anxiety causes us to work faster, and when we work faster, we’re often unable to recognize patterns that would be clearer to us if we were calm.

There’s interesting research on this. Psychologists, knowing that the color red prompts an anxiety response and that the color blue has a calming effect, conducted a study in which test-takers had to answer math questions – the questions were given to some subjects on paper with a red background and to other subjects on paper with a blue background. (The control group had questions on standard white paper.) The red anxiety-producing background noticeably lowered scores and the calming blue background boosted scores.

Now, the GMAT doesn’t give you a red background, but it does give you unfriendly-seeming numbers that likely have the same effect. So, this question is as much about psychology as it is about mathematical proficiency. Our job is to take a deep breath or two and rein in our anxiety before we proceed.

If Dara has completed 10% of her workout, we know she has 90% of her workout remaining. So, that 24 min. 18 sec. presents 90% of her total workout. If we designate her total workout time as “t,” we end up with the following equation:

24 min. 18 sec. = 0.90t

Let’s work with fractions to solve. 18 seconds is 18/60 minutes, which simplifies to 3/10 minutes. 0.9 is 9/10, so we can rewrite our equation as:

24 + 3/10 = (9/10)t
(243/10) = (9/10)t
(243/10)*(10/9) = t
27 = t

Not so bad. Dara’s full workout is 27 minutes long.

We want to know how much time is remaining when Dara has completed 40% of her workout. Well, if she’s completed 40% of her workout, we know she has 60% of her workout remaining. If her full workout is 27 minutes, then 60% of this value is 0.60*27 = (3/5)*27 = 81/5 = 16 + 1/5, or 16 minutes 12 seconds. And we’ve got our answer: E.

Now, let’s say you get this problem with 20 seconds remaining on the clock and you simply don’t have time to solve it properly. Let’s estimate.

We know that the correct answer is over 16 minutes and that we’ve significantly underestimated – makes sense to go with E.

Takeaway: Don’t let the question-writer trip you up with figures concocted to make you nervous. Take a breath, and remember that the concepts being tested are the same ones that, when boiled down to their essence, are a breeze when we’re calm.